## RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.1
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.2
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.3
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.4
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.5
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.6
- RD Sharma Class 10 Solutions Chapter 7 Triangles Ex 7.7
- RD Sharma Class 10 Solutions Chapter 7 Triangles Revision Exercise
- RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS
- RD Sharma Class 10 Solutions Chapter 7 Triangles MCQS

**Answer each of the following questions either in one word or one sentence or as per requirement of the questions :**

**Question 1.**

State basic proportionality theorem and its converse.

**Solution:**

**Basic Proportionality Theorem :** If a line is drawn parallel to one side of a triangle intersects the other two sides in distinct points, then the other two sides are divided in the same ratio.

**Conversely :** In a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.

**Question 2.**

In the adjoining figure, find AC.

**Solution:**

**Question 3.**

In the adjoining figure, if AD is the bisector of ∠A, what is AC ?

**Solution:**

In the figure, AD is the angle bisector of ∠A of ∆ABC

AB = 6 cm, BC = 3 cm, DC = 2 cm

**Question 4.**

State AAA similarity criterion.

**Solution:**

If in two triangles, corresponding angles are respectively equal then the triangles are similar.

**Question 5.**

State SSS similarity criterion.

**Solution:**

If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar.

**Question 6.**

State SAS similarity criterion.

**Solution:**

If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.

**Question 7.**

In the adjoining figure, DE is parallel to BC and AD = 1 cm, BD = 2 cm. What is the ratio of the area of ∆ABC to the area of ∆ADE?

**Solution:**

In ∆ABC, DE || BC

**Question 8.**

In the figure given below, DE || BC. If AD = 2.4 cm, DB = 3.6 cm and AC = 5 cm. Find AE.

**Solution:**

In ∆ABC, DE || BC

AD = 2.4 cm, DB = 3.6 cm, AC = 5 cm

x = 2

AE = 2 cm

**Question 9.**

If the areas of two similar triangles ABC and PQR are in the ratio 9 : 16 and BC = 4.5 cm, what is the length of QR ?

**Solution:**

∆ABC ~ ∆PQR

Area of ∆ABC : area of ∆PQR = 9 : 16

BC = 4.5 cm

Let QR = A

The area of two similar triangles are in the ratio of the squares of their corresponding sides

**Question 10.**

The areas of two similar triangles are 169 cm² and 121 cm² respectively. If the longest side of the larger triangle is 26 cm, what is the length of the longest side of the smaller triangle ?

**Solution:**

Let ∆ABC be the larger triangle and ∆PQR

be the smaller triangle and their longest sides be BC and QR respectively

Area of ∆ABC = 169 cm²

area of ∆PQR = 121 cm²

BC = 26 cm

Let QR = x cm

∆ABC ~ ∆PQR

**Question 11.**

If ABC and DEF are similar triangles such that ∠A = 57° and ∠E = 73°, what is the measure of ∠C ?

**Solution:**

∆ABC ~ ∆DEF

Their corresponding angles are equal ∠A = ∠D, ∠B = ∠E and ∠C = ∠F

But ∠A = 57°

**Question 12.**

If the altitude of two similar triangles are in the ratio 2 : 3, what is the ratio of their areas ?

**Solution:**

Let ∆ABC ~ ∆PQR

and let AL ⊥ BC and PM ⊥ QR

**Question 13.**

**Solution:**

**Question 14.**

If ∆ABC and ∆DEF are similar triangles such that AB = 3 cm, BC = 2 cm, CA = 2.5 cm and EF = 4 cm, write the perimeter of ∆DEF.

**Solution:**

**Question 15.**

State Pythagoras Theorem and its converse.

**Solution:**

In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Conversely : If in a triangle, the square of one side is equal to the sum of squares of the remaining two sides, then the angle opposite to the first side is a right angle.

**Question 16.**

The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus. **(C.B.S.E. 2008)**

**Solution:**

In rhombus ABCD, BD = 30 cm, AC = 40 cm

**Question 17.**

In figure, PQ || BC and AP : PB = 1 : 2.

**Solution:**

In ∆ABC, PQ || BC

∆APQ ~ ∆ABC

But AP : PB = 1 : 2

The ratio of the areas of two similar triangles are proportional to the square of their corresponding sides

**Question 18.**

In the figure, S and T are points on the sides PQ and PR respectively of ∆PQR such that PT = 2 cm, TR = 4 cm and ST is parallel to QR. Find the ratio of the areas of ∆PST and ∆PQR. **[CBSE 2010]**

**Solution:**

In ∆PQR, ST || QR and PT = 2 cm, TR = 4 cm

PR = PT + TR = 2 + 4 = 6 cm

**Question 19.**

In the figure, ∆AHK is similar to ∆ABC. If AK = 10 cm, BC = 3.5 cm and HK = 7 cm, find AC. **[CBSE 2010]**

**Solution:**

In the given figure, AK = 10 cm, BC = 3.5 cm, HK = 7 cm

**Question 20.**

In the figure, DE || BC in ∆ABC such that BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.

**Solution:**

In the given figure,

DE || BC

BC = 8 cm, AB = 6 cm and DA = 1.5 cm

**Question 21.**

In the figure, DE || BC and AD = \(\frac { 1 }{ 2 }\) BD. If BC = 4.5 cm, find DE. **[CBSE 2010]**

**Solution:**

In the given figure,

DE || BC

**Question 22.**

In the figure, ∠M = ∠N = 46°. Express x in terms of a, b and c where a, b, c are lengths of LM, MN and NK respectively.

**Solution:**

In the figure, ∠M = ∠N = 46°

∠M = a, PN = x, MN = b, NK = c

∠M = ∠N = 46°

But there are corresponding angle

PN || ML

∆PKN ~ ∆LKM

Hope given RD Sharma Class 10 Solutions Chapter 7 Triangles VSAQS are helpful to complete your math homework.

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