NCERT Class 10 Math Chapter 6 Important Questions Answer – Triangles

Class 10
Chapter  Triangles
Subject Math
Category Important Question Answer

Class 10 Math Chapter 6 Important Question Answer

Q1. Two polygons of the same number of sides are similar, if their corresponding angles are ___________. (equal, proportional)

Ans. Equal

Q2. If areas of two similar triangles are equal, then prove that they are congruent.


Let us take ∆ ABC ~ ∆ DEF. As we know in similar triangles, corresponding angles and corresponding sides are equal. So, ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and corresponding sides ratios

 \displaystyle \frac{{AB}}{{DE}} =  \displaystyle \frac{{BC}}{{EF}} =  \displaystyle \frac{{CA}}{{FD}}

We know the theorem that if two triangles are similar, then the ration of the area of both triangles is proportional to the square of the ration of their corresponding sides. So we have

 \displaystyle \frac{{\text{Area of }!!\Delta!!\text{ ABC}}}{{\text{Area of }!!\Delta!!\text{ DEF}}} =  \displaystyle {{\left( {\frac{{AB}}{{DE}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{BC}}{{EF}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{CA}}{{FD}}} \right)}^{2}}

We are given in the question that Area of ∆ ABC = Area of ∆ DEF. So we have,

 \displaystyle \frac{{\text{Area of }!!\Delta!!\text{ ABC}}}{{\text{Area of }!!\Delta!!\text{ DEF}}} =  \displaystyle {{\left( {\frac{{AB}}{{DE}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{BC}}{{EF}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{CA}}{{FD}}} \right)}^{2}}

1 =  \displaystyle {{\left( {\frac{{AB}}{{DE}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{BC}}{{EF}}} \right)}^{2}} =  \displaystyle {{\left( {\frac{{CA}}{{FD}}} \right)}^{2}}

On solving we get,

1 =  \displaystyle {{\left( {\frac{{AB}}{{DE}}} \right)}^{2}}

Square rooting both side, we get

1 =  \displaystyle \frac{{AB}}{{DE}}

AB = DE,

Similarly, 1 =  \displaystyle {{\left( {\frac{{BC}}{{EF}}} \right)}^{2}} i.e BC = EF

And 1 =  \displaystyle {{\left( {\frac{{CA}}{{FD}}} \right)}^{2}} i.e CA = FD

Since the length of the corresponding sides of the triangles are equal by side-side-side criteria the triangles are congruent.

∆ ABC ≅ ∆ DEF

Hence proved.

Q3. All squares are ___________. (similar, congruent) Most Important

Ans. Similar.

Q4. All  __________ triangles are similar. (isosceles, equilateral) Most Important

Ans. equilateral

Q5. Sides of triangles are given below. Determine which of them is a right triangle (i) 3 cm, 8 cm, 6 cm (ii) 13 cm, 12 cm, 5 cm

Ans. In a right-angle triangle, the square of hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras Theorem). i.e h2 = p2 + b2, where h is the longest side of triangle.

  • h2 = 82 = 64

p2 + b2 = 32 + 62 = 45

As, 64 ≠ 45

Therefore, it is not a right-angle triangle.

  • h2 = 132 = 169

p2 + b2 = 122 + 52 = 169

169 = 169

Therefore, triangle with sides 13 cm, 12 cm, 5 cm is a right-angle triangle.

Q6. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.


Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp post.

 From the fig, DE is the shadow of the girl. Let DE be x metres.

Now, BD = 1.2m × 4 = 4.8m

In ∆ ABE and ∆ CDE,

∠B = ∠D (Each is of 90° because lamp-post as well as the girl is standing vertical to the ground)

∠E = ∠E (Same angle)

So, ∆ ABE ~ ∆ CDE (AA similarity criterion)


 \displaystyle \frac{{BE}}{{DE}} =  \displaystyle \frac{{AB}}{{CD}}

 \displaystyle \frac{{4.8+x}}{x} =  \displaystyle \frac{{3.6}}{{0.9}}         ( 90cm =  \displaystyle \frac{{90}}{{100}}m = 0.9m)

4.8 + x = 4x

3x = 4.8

x = 1.6

So, the shadow of the girl after walking for 4 seconds is 1.6m long.

Q7. In figure DE || BC. Find the value of EC: Most Most Important

Ans. Using basic proportionality theorem,  \displaystyle \frac{{AD}}{{DB}} =  \displaystyle \frac{{AE}}{{EC}}

 \displaystyle \frac{2}{4} =  \displaystyle \frac{1}{EC}

EC = 2 cm

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