## Triangles Class 10 MCQ Questions with Answers

Question 1.

## ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Answer:

(C) 4 :1

Explanation:

\(\frac{a T \triangle A B C}{a r \triangle B D E}\) =\(\frac{\frac{\sqrt{3}}{4}(B C)^{2}}{\frac{\sqrt{3}}{4}(B D)^{2}}\) = \(\frac{(B C)^{2}}{(B D)^{2}}\)

= \(\frac{(B C)^{2}}{\left(\frac{B C}{2}\right)^{2}}\) = \(\frac{4 B C^{2}}{B C^{2}}\) = \(\frac{4}{1}\) = 4 : 1

Question 2.

## Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio :

(A) 2 : 3

(B) 4 : 9

(C) 81:16

(D) 16 : 81

Answer:

(D) 16 : 81

Explanation:

We know that the ratio of the areas of the triangles will be equal to the square of the ratio of the corresponding sides of the triangles.

Thus, required ratio of the areas of the two

triangles = \(\left(\frac{4}{9}\right)^{2}\) = \(\frac{16}{81}\)

Question 3.

## In the figure given below, ∠BAC = 90° and AD 1 BC. Then:

(A) BD x CD = BC^{2}

(B) AB x AC = BC^{2}

(C) BD x CD = AD^{2}

(D) AB x AC = AD^{2}

Answer:

(C) BD x CD = AD^{2
}

Explanation:

In ∆ABD and ∆CAD, ∠ADB = √ADC = 90° and ∠ABD = ∠CAD = θ By AA Similarity, we get, ∆ABD ~ ∆CAD

⇒ \(\vec{a}\) = \(\vec{a}\) ⇒ BD x CD = AD^{2}

Question 4.

## If ∆ABC ~ ∆EDF and AABC is not similar to ∆DEF, then which of the following is not true?

(A) BC x EF = AC x FD

(B) AB x EF = AC x DE

(C) BC x DE = AB x EF

(D) BC x DE = AB x FD

Answer:

(C) BC x DE = AB x EF

Explanation:

Since ∆ABC ~ ∆EDF, then we get AB = \(\frac{A B}{E D}\) = \(\frac{A C}{E F}\) = \(\frac{B C}{D F}\)

From first two, AB x EF = AC x DE.

Option (B) is correct.

From last two, BC x EF = AC x FD.

Option (A) is correct.

From first and last, BC x DE = AB x FD.

Option (D) is correct.

Thus, option (C) is incorrect

Question 5.

## If in two triangles ABC and PQR, \(\frac{A B}{Q R}\) = \(\frac{B C}{P R}\) = \(\frac{C A}{P Q}\)

then:

(A) ∆PQR – ∆CAB

(B) ∆PQR – ∆ABC

(C) ∆CBA – ∆PQR

(D) ∆BCA – ∆PQR

Answer:

(A) ∆PQR – ∆CAB

Explanation: Given that, \(\frac{A B}{Q R}\) =\(\frac{B C}{P R}\) = \(\frac{C A}{P Q}\) , by SSS similarity, we get ∆PQR ~ ∆CAB.

Question 6.

## In the figure given below/- two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC 2.5 cm, PD = 5 cm, ∠APB = 50° and ∠CDP = 30°. Then, ∠PBA is equal to :

(A) 50°

(B) 30°

(C) 60°

(D) 100°

Answer:

(D) 100°

Explanation:

In the given figure, \(\frac{P A}{P B}\) = \(\frac{6}{3}\) =2

and \(\frac{P D}{P C}\) = \(\frac{5}{2.5}\) = 2 Thus \(\frac{P A}{P B}\) = \(\frac{P D}{P C}\)

∠APB = ∠DPC. By SAS similarity, we get ∆APB ~ ∆DPC. Hence,∠PBA = 100°

Question 7.

## If in two triangles DEF and PQR, ∠D = ∠Q and ∠R = ∠E, then which of the following is not true?

(A) \(\frac{E F}{P R}\) = \(\frac{D F}{P Q}\)

(B) \(\frac{D E}{P Q}\) = \(\frac{F E}{R P}\)

(C) \(\frac{D E}{Q R}\) = \(\frac{D F}{P Q}\)

(D) \(\frac{E F}{R P}\) = \(\frac{D E}{Q R}\)

Answer:

(B) \(\frac{D E}{P Q}\) = \(\frac{F E}{R P}\)

Explanation:

In ∆DEF and ∆PQR, ∠D – ∠Q and ∠R – ∠E. By AA similarity, wo get ∆DEE ~ ∆QRP. Hence, \(\frac{D E}{Q R}\) = \(\frac{E F}{R P}\) = \(\frac{D F}{Q P}\) . Option (B) is incorrect.

Question 8.

## In triangles ABC and DEF, ∠B = ∠E, ∠E = ∠C and AB = 3DE. Then, the two triangles are:

(A) congruent but not similar

(B) similar but not congruent

(C) neither congruent nor similar

(D) congruent as well as similar

Answer:

(B) similar but not congruent

Explanation:

In ∆ABC and ∆DEE, ∠B = ∠E and ∠E – ∠C. By AA similarity, we get ∆ABC ~ ∆DEE. Thus, the triangles arc similar but not congruent.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.

(A) Both A and R are true and R is the correct explanation of A

(B) Both A and R are true but R is NOT the correct explanation of A

(C) A is true but R is false

(D) A is false and R is True

Question 1.

Assertion (A): If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, then triangles will be similar.

Reason (R): If the ratio of the corresponding altitudes of two similar triangles is \(\frac{3}{5}\) ratio of their areas is\(\frac{6}{5}\).

Answer;

(C) A is true but R is false

Explanation:

In case of assertion:

In the given two right triangles, both have equal right angles and one of the acute angles of one triangle is equal to an acute angle of the other triangle.

Thus, by AA similarity, the given two triangles are similar.

∴ Assertion is correct.

In case of reason:

We know that the ratio of the areas of two similar triangles is the square of the ratio of the corresponding altitudes of two similar triangles.

Thus, the ratio of the areas of two similar triangles is – \(\left(\frac{3}{5}\right)^{2}\) = \(\frac{9}{25}\)

∴Reason is incorrect.

Hence, assertion is correct but reason is incorrect.

Question 2.

## Assertion (A): If D is a point on side QR of ∆PQR such that PD ⊥QR, then ∆PQD ~ ∆RPD.

Reason (R): In the figure given below, if ∠D = ∠C then ∆ADE ~ ∆ACB.

Answer:

(D) A is false and R is True

Explanation:

In case of assertion:

IN ∆PQD and ∠PDQ = ∠PDR = 90°

There is no other information to be similar. Thus, is not it will be correct to say that AAPQD = ARPD.

∴ Assertion is correct.

In case of assertion:

IN AADE and AACB, we have

∠ADE = ∠ACB [given ]

∠DAE = ∠CAB [Common angle]

By AA similarity, we get ∆ADE ~ ∆ACB

∴ Reason is correct

Hence, assertion is incorrect but reason is correct

Question 3.

## Assertion (A): Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm^{2}, then the area of the larger triangle is 108 cm^{2}.

Reason (R): If D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC, then CA^{2} = CB x CD.

Answer:

(B) similar but not congruent

Explanation:

In case of assertion:

Let ∆ABC and ∆DEE are two similar triangles.

\(\frac{a r \triangle A B C}{a r \triangle D E F}\) = \(\left(\frac{A B}{D E}\right)^{2}\)

Given that ar ∆ABC = 48 cm^{2}. Then,

\(\frac{48}{a r \Delta D E F}\)

= \(\frac{4}{9}\)

⇒ ar ∆DEF = \(\vec{a}\) x 48 = 108

Thus, the area of larger triangle is 108 cm^{2}.

∴Assertion is correct.

In case of reason:

In ∆BAC and ∆ADC,

∠BAC = ∠ADC [Given]

∠BCA = ∠ACD [Common angle]

By AA similarity, ∆BAC ~ ∆ADC Thus,

\(\frac{C A}{C D}\) = \(\frac{B C}{C A}\)

∴ Reason is correct

Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 4.

## Assertion (A): In an equilateral triangle of side 3√3 cm, then the length of the altitude is 4.5 cm.

Reason (R): If a ladder 10 cm long reaches a window 8 m above the ground, then the distance of the foot of the ladder from the base of the wall is 6 m.

Answer:

(B) similar but not congruent

Explanation:

In case of assertion:

∆ABD, ∠D = 90°

∴(3√3)^{2} = h^{2} + \(\left(\frac{3 \sqrt{3}}{2}\right)^{2}\)

or, 27 = h2 + \(\vec{a}\)

or,h2 = 27 – \(\vec{a}\)

or, h2 = \(\vec{a}\)

∴ h = \(\vec{a}\) = 4.5m

∴ Assertion is correct.

In case of assertion:

Let BC be the height of the window above the ground and AAAC be a ladder.

Here, BC = 8cm and AC = 10 cm

∴ In right angled triangle ABC,

AC^{2} = AB^{2} + BC^{2}(By using Pythagoras Theorem)

(10)2 = AB2 + (8)2

AB2 = 100 – 64

= 36

AB = 6 m.

∴ Reason is correct

Hence, both assertion and reason are correct but reason is not the correct explanation for as sertion.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.

I. Read the following text and answer the following questions on the basis of the same: SCALE FACTOR AND SIMILARITY SCALE FACTOR A scale drawing of an object is of the same shape as the object but of a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents.The value of scale is written as a ratio.SIMILAR FIGURES The ratio of two corresponding sides in similar fig-ures is called the scale factor.

Scale factor = \(\vec{a}\)

If one shape can become another using Resizing then the shapes are Similar.

Hence, two shapes aaaarae Similar when one can become the other after a reisze, flip, slide or turn.

Question 1.

## A model of a boat is made on the scale of 1 : 4. The model is 120 cm long. The full size of the boat has a width of 60 cm. What is the width of the scale model ?

(A) 20 cm

(B) 25 cm

(C) 15 cm

(D) 240 cm

Answer:

(C) 15 cm

Explanation:

Width of the scale model = 60/4 = 15 cm.

Question 2.

## What will effect the similarity of any two polygons ?

(A) They are flipped horizontally

(B) They are dilated by a scale factor

(C) They are translated down

(D) They are not the mirror image of on another.

Answer:

(D) They are not the mirror image of on another.

Explanation:

They are not the mirror image of one another.

Question 3.

## If two similar triangles have a scale factor of a : b, which statement regarding the two triangles is true ?

(A) The ratio of their perimeters is 3 a:b

(B) Their altitudes have a ratio a : b

(C) Their medians have a ratio \(\frac{a}{2}\): b

(D) Their angle bisectors have a ratio a^{2} : b^{2}

Answer:

(B) Their altitudes have a ratio a : b

Explanation:

Let ABC and PQR be two simliar triangles and AD,PE are two altitudes:

\(\frac{A B}{P Q}\) = \(\frac{B C}{Q R}\) = \(\frac{A C}{P R}\)

(corresponding sides)

∠B = ∠Q and ∠ADB = ∠PEQ (each 90°)

Now, \(\frac{A D}{P E}\) = \(\frac{A B}{P Q}\) = \(\frac{a}{b}\)

Question 4.

## The shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is

(A) 3 m

(B) 3.5 m

(C) 4. 5 m

(D) 5 m

Answer:

(D) 5 m

Explanation:

Let shadow of the tree be x. By the property to similar triangles we have \(\frac{5}{2}\) = \(\frac{12.5}{x}\)

x = \(\frac{(12.5 \times 2)}{5}\) = 5 m

Question 5.

## Below you see a student’s mathematical model of a farmhouse roof with measurements. The attic floor, ABCD in the model, is a square. The beams that support the roof are the edge of a rectangular prism, EFGHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT, and H is the middle of DT. All the edges of the pyramid in the model have length of 12 m.

What is the length of EF, where EF is one of the horizontal edges of the block ?

(A) 24 m

(B) 3 m

(C) 6 m

(D) 10 m

Answer:

(C) 6 m

Explanation:

Length of the horizontal edge EF = half of the edge of pyramid =\(\vec{a}\) = 6 cm (as E is he mid-point of AT)

II. Read the following text and answer the below questions:

Seema placed a light bulb at point O on the ceiling and directly below it placed a table. Now, she put a cardboard of shape ABCD between table and lighted bulb. Then a shadow of ABCD is casted on the table as A’B’C’D’ (see figure). Quadrilateral A’B’C’D’ in an enlargement of ABCD with scale factor 1 : 2, Also, AB = 1.5 cm, BC = 25 cm, CD = 2.4 cm and AD = 2.1 cm; ∠A = 105°, ∠B = 100°, ∠C = 70° and ∠D = 85°.

Question 1.

## What is the measurement of angle A?

(A) 105°

(B) 100°

(C) 70°

(D) 80°

Answer:

(A) 105°

Explanation:

Quadrilateral A’B’C’D’ is similar to ABCD.

∴∠A = ∠A

⇒∠A = 105°

Question 2.

## What is the length of A’B’ ?

(A) 1.5 cm

(B) 3 cm

(C) 5 cm

(D) 2.5 cm

Answer:

(B) 3 cm

Explanation:

Given scale factor is 1 : 2

∴ A’B’ = 2AB

⇒ A’B’ = 2 x 1.5 = 3 cm

Question 3.

## What is the sum of angles of quadrilateral A’B’C’D’ ?

(A) 180°

(B) 360°

(C) 270°

(D) None of these

Answer:

(B) 360°

Explanation:

Sum of the angles of quadrilateral A’B’C’D’ is 360°

Question 4.

## What is the ratio of sides A’B’ and A’D’ ?

(A) 5 : 7

(B) 7 : 5

(C) 1 : 1

(D) 1 : 2

Answer:

(A) 5 : 7

Explanation:

A’B’ = 3 cm

and A’D’ = 2AD

= 2 x 2.1 = 4.2 cm

\(\frac{A^{\prime} B^{\prime}}{A^{\prime} D^{\prime}}\) = \(\frac{3}{4.2}\) = \(\frac{30}{42}\)

= \(\frac{5}{7}\) or 5 : 7

Question 5.

## What is the sum of angles C’ and D’ ?

(A) 105°

(B) 100°

(C) 155°

(D) 140°

Answer:

(C) 155°

Explanation:

∠C = ∠C = 70°

and ∠D’ = ∠D = 85°

∴∠C’ + ∠D’ = 70° + 85° = 155°

III. Read the following text and answer the below questions: SIMILAR TRIANGLES Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles. The height of Vijay’s house if 20 m when Vijay’s house casts a shadow 10 m long on the ground. At the same time, the tower casts a shadow 50 m long on the ground and the house of Ajay casts 20 m shadow on the ground.

Question 1.

## What is the height of the tower?

(A) 20 m

(B) 50 m

(C) 100 m

(D) 200 m

Answer:

(C) 100 m

Explanation:

When two corresponding angles of two triangles are simliar, then ratio of sides are equal

\(\frac{\text { Height of Vjay’s house }}{\text { Length of Shadow }}\)

= \(\frac{\text { Height of tower }}{\text { Length of Shadow }}\)

\(\frac{20 \mathrm{~m}}{10 \mathrm{~m}}\) = \(\frac{\text { Height of tower }}{\text { 50 m }}\)

Height of tower = \(\frac{20 \times 50}{10}

\) = \(\frac{1000}{10}

\) = 100 m

Question 2.

## What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12 m?

(A) 75 m

(B) 50 m

(C) 45 m

(D) 60 m

Answer:

(D) 60 m

Question 3.

## What is the height of Ajay’s house?

(A) 30 m

(B) 40 m

(C) 50 m

(D) 20 m

Answer:

(B) 40 m

Explanation:

\(\frac{\text { Height of Vijay’s house}}{\text { Length of shadow }}\)

=\(\frac{\text { Height of Aijay’s house}}{\text { Length of shadow }}\)

\(\frac{20 \mathrm{~m}}{10 \mathrm{~m}}\) = \([latex]\frac{\text { Height of Aijay’s house}}{\text { 20 m }}\)

Height of Ajay’s house = \(\frac{20 \mathrm{~m} \times 20 \mathrm{~m}}{10 \mathrm{~m}}

\)

= 40 m.

Question 4.

## When the tower casts a shadow of 40 m, same time what will be the length of the shadow of Ajay’s house?

(A) 16 m

(B) 32 m

(C) 20 m

(D) 8 m

Ans.

(A) 16 m

Question 5.

## When the tower casts a shadow of 40 m, same time what will be the length of the shadow of Vijay’s house?

(A) 15 m

(B) 32 m

(C) 16 m

(D) 8 m

Answer:

(D) 8 m

IV. Read the following text and answer the below questions:

Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with another point C are a distance of 12 m, connecting C to point D at a distance of 40 m from point C and the connecting D to the point A which is are a distance of 30 m from D such the ∠ADC = 90°

Question 1.

## Which property of geometry will be used to find the distance AC?

(A) Similarity of triangles

(B) Thales Theorem

(C) Pythagoras Theorem

(D) Area of similar triangles

Answer:

(C) Pythagoras Theorem

Question 2.

## What is the distance AC?

(A) 50 m

(B) 12 m

(C) 100 m

(D) 70 m

Answer:

(A) 50 m

Explanation:

According to the pythagoras.

AC^{2} = AD^{2}+ CD^{2}

AC^{2} = (30 m)^{2} + (40 m)^{2}

AC^{2} = 900 + 1600

AC^{2} = 2500

AC = 50 m

Question 3.

## Which is the following does not form a Pythagoras triplet?

(A) (7,24,25)

(B) (15,8,17)

(C) (5,12,13)

(D) (21,20,28)

Answer:

(D) (21,20,28)

Question 4.

## Find the length AB?

(A) 12 m

(B) 38 m

(C) 50 m

(D) 100m

Answer:

(B) 38 m

Question 5.

## Find the length of the rope used.

(A) 120 m

(B) 70 m

(C) 82 m

(D) 22 m

Ans.

(C) 82 m

Explanation:

Length of Rope = BC + CD+DA

= 12 m + 40 m + 30 m

= 82 m

V. Read the following text and answer the below questions: SCALE FACTOR A scale drawing of an object is the same shape at the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor Scale factor = length in image / corresponding length in object

If one shape can become another using revising, then the shapes are similar. Hence, two shapes are similar when one can become the other after a resize, flip, slide or turn. In the photograph below showing the side view of a train engine. Scale factor is 1:200 This means that a length of 1 cm on the photograph above corresponds to a length of 200 cm or 2 m, of the actual engine. The scale can also be written as the ratio of two lengths.

Question 1.

## If the length of the model is 11 cm, then the overall length of the engine in the photograph above, including the couplings(mechanism used to connect) is:

(A) 22 cm

(B) 220 cm

(C) 220 m

(D) 22 m

Answer:

(A) 22 cm

Question 2.

## What will affect the similarity of any two polygons?

(A) They are flipped horizontally

(B) They are dilated by a scale factor

(C) They are translated down

(D) They are not the mirror image of one another.

Answer:

(D) They are not the mirror image of one another.

Question 3.

## What is the actual width of the door if the width of the door in photograph is 0.35 cm?

(A) 0.7 m

(B) 0.7 cm

(C) 0.07 cm

(D) 0.07 m

Answer:

(A) 0.7 m

Question 4.

## If two similar triangles have a scale factor 5:3 which statement regarding the two triangles is true?

(A) The ratio of their perimeters is 15:1

(B) Their altitudes have a ratio 25:15

(C) Their medians have a ratio 10:4

(D) Their angle bisectors have a ratio 11:5

Answer:

(B) Their altitudes have a ratio 25:15

Question 5.

## The length of AB in the given figure: A

(A) 8 cm

(B) 6 cm

(C) 4 cm

(D) 10 cm

Answer:

(C) 4 cm

Explanation:

Since AABC and AADE are similar, then their ratio of corresponding sides are equal

\(\frac{A B}{B C}\) = \(\frac{A B+B D}{D E}\)

\(\frac{x}{3 \mathrm{~cm}}\) = \(\frac{(x+4) \mathrm{cm}}{6 \mathrm{~cm}}\)

6x = 3(x + 4)

6x = 3x + 12

6x – 3x = 12

3x = 12

x = 4

AB = 4 cm.