- Similar Figures
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- Similar Polygons
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- Basic Proportionally Theorem (or Thales Theorem)
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- Criteria for Similarity of Triangles
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- Different Criterion for similarity of the triangles
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- Areas of Similar Triangles
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- Pythagoras Theorem
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- Converse of Pythagoras Theorem

ABC is an isosceles triangle, right -angled at C. Prove that $AB^2 = 2BC^2$.

Now Since it is an isosceles triangle

AC=BC -(1)

Now from Pythagorean theorem

$AB^2 = AC^2 + BC^2$

From

$AB^2 = 2BC^2$

The areas of two similar triangles ABC and PQR are in the ratio 9 : 16. If BC = 4.5 cm, find the length of QR.

We know that

$\frac {ar( \Delta ABC)}{ar( \Delta PQR)}=\frac {BC^2}{QR^2}$

$\frac {9}{16} = \frac {4.5^2}{QR^2}$

$QR= 6 cm$

The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its tip reach?

First case is depicted in the figure (a)

Now from Pythagoras theorem

$AC^2 = AB^2 + BC^2$

$AC^2 = 8^2 + 6^2 =100$

$AC=10$ cm

This is the length of the ladder.

Now Second case is depicted in figure (b)

AC=10 cm

BC=8 cm

Now from Pythagoras theorem

$AC^2 = AB^2 + BC^2$

$AB^2 = AC^2 - BC^2=10^2 - 8^2 =36$

$AB=6$ cm

Determine whether the triangle having sides $(b - 1)$ cm, $2\sqrt {b}$ cm and $(b + 1)$ cm is a right angled triangle.

These are sides of the Right angle triangle

$(b+1)^2 = (2\sqrt {b})^2 + (b-1)^2$

$(b+1)^2 = 4b + b^2 +1 -2b$

$(b+1)^2=b^2 + 1 +2 b$

$(b+1)^2 = (b+1)^2$

The areas of two similar triangles are 121 cm

We know that

$\frac {ar( \Delta 1)}{ar( \Delta 2)}=\frac {M_1^2}{M_2^2}$

$\frac {121}{64} = \frac {12.1^2}{QR}$

$QR= 8.8 cm$

In $ \Delta ABC$, AD is perpendicular to BC. Prove that:

a. $AB^2+ CD^2 = AC^2 + BD^2$

b. $AB^2 - BD^2 = AC^2 - CD^2$

In $ \Delta ABC$, AD is perpendicular to BC

Now $ \Delta ABD$ is an right -angle triangle

So from Pythagoras theorem

$AB^2 = AD^2 + BD^2$

$AD^2 = AB^2 - BD^2$ -(1)

Similarly $ \Delta ADC$ is an right -angle triangle

So from Pythagoras theorem

$AC^2 = AD^2 + CD^2$

$AD^2 = AC^2 - CD^2$ --(2)

From (1) and (2)

$AB^2 - BD^2 = AC^2 - CD^2$

Which proved part (b)

Now rearranging,

$AB^2+ CD^2 = AC^2 + BD^2$

Which proved part (a)

In a quadrilateral ABCD, given that $\angle A + \angle D = 90^0$. Prove that $AC^2 + BD^2 = AD^2 + BC^2$

ABCD is a quadrilateral as shown below. Extending the side CD and AB to meet at point P

Now in $\Delta APD$

$\angle A + \angle D + \angle C =180$

Now $\angle A + \angle D = 90^0$

Therefore

$\angle C=90$

Now in right angle triangle APC

$AC^2 = AP^2 + PC^2$

Now in right angle triangle BPD

$BD^2 = BP^2 + DP^2$

Adding both

$AC^2 + BD^2 =AP^2 + PC^2 + BP^2 + DP^2$ -(1)

Now in right angle triangle APD

$AD^2 = AP^2 + DP^2$

Now in right angle triangle BCP

$BC^2 = CP^2 + BP^2$

Adding both

$AD^2 + BC^2 = AP^2 + PC^2 + BP^2 + DP^2$ -(2)

From (1) and (2)

$AC^2 + BD^2 = AD^2 + BC^2$

Prove that a line drawn through the mid- point of one side of a triangle parallel to another side bisects the third side.

Prove that the line joining the mid – points of any two sides of a triangle is parallel to the third side

Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm,4 cm,5 cm

(iii) 40 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

i. $25^2 = 7^2 + 24^2$

Hence right angle triangle with 25 cm as hypotenuse

ii. $5^2 = 3^2 + 4^2$

Hence right angle triangle with 5 cm as hypotenuse

iii. $100^2 = 40^2 + 80^2$

Hence right angle triangle with 100 cm as hypotenuse

iv. $13^2 = 12^2 + 5^2$

Hence right angle triangle with 13 cm as hypotenuse

DEF is an equilateral triangle of side $2b$. Find each of its altitudes.

In right angle triangle DNF

$DF^2 = DN^2 + NF^2$

$DN^2 = DF^2 - NF^2$

$DN^2 = 4b^2 - b^2$

$DN =b \sqrt {3}$

Triangle ABC is right- angled at B and D is the mid - point of BC.

Prove that: $AC^2 = 4AD^2 - 3AB^2$

In right angle triangle ABC

$AC^2 = AB^2 + BC^2$ --(1)

In right triangle ABD

$AD^2 = AB^2 + BD^2$

$AD^2 =AB^2 + \frac {BC^2}{4}$

$BC^2 = 4AD^2 - 4 AB^2$

Substituting this in equation (1)

$AC^2 = 4AD^2 - 3AB^2$

The sides of a triangle ABC are in the ratio AB : BC : CA = 1 : √2: 1. Show that ABC is a right triangle, right -angled at A.

AB : BC : CA = 1 : √2: 1

$AB=x$

$BC= x\sqrt {2}$

$CA=x$

Now

$BC^2 = AB^2 + AC^2$

$2x^2 = x^2 +x^2$

$2x^2=2x^2$

So ABC is right -angled at A

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower

Given

AB = 6m

BC= 4 m

Let Height of tower (DE)=h m

EF = 28 m

In $\Delta ABC$ and $\Delta DEF

$

$\angle C = \angle F$ (Same angle of elevation of Sun)

$ \angle B = \angle E = 90$

By AA similarity criterion

$\Delta ABC \sim \Delta DEF$

$\frac {AB}{DE}=\frac {BC}{EF}$

$ h=42 m$

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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