Question 1. ABC is an isosceles triangle, right -angled at C. Prove that $AB^2 = 2BC^2$. Solution
Now Since it is an isosceles triangle
AC=BC -(1)
Now from Pythagorean theorem
$AB^2 = AC^2 + BC^2$
From
$AB^2 = 2BC^2$
Question 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. Solution
We know that
$\frac {ar( \Delta ABC)}{ar( \Delta PQR)}=\frac {BC^2}{QR^2}$
$\frac {9}{16} = \frac {4.5^2}{QR^2}$
$QR= 6 cm$
Question 3. 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? Solution
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
Question 4. Determine whether the triangle having sides $(b - 1)$ cm, $2\sqrt {b}$ cm and $(b + 1)$ cm is a right angled triangle. Solution
These are sides of the Right angle triangle
$(b+1)^2 = (2\sqrt {b})^2 + (b-1)^2$
$(b+1)^2 = 4b + b^2 +1 -2b$
$(b+1)^2=b^2 + 1 +2 b$
$(b+1)^2 = (b+1)^2$
Question 5. The areas of two similar triangles are 121 cm^{2} and 64 cm^{2} respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other. Solution
We know that
$\frac {ar( \Delta 1)}{ar( \Delta 2)}=\frac {M_1^2}{M_2^2}$
$\frac {121}{64} = \frac {12.1^2}{QR}$
$QR= 8.8 cm$
Question 6. In $ \Delta ABC$, AD is perpendicular to BC. Prove that:
a. $AB^2+ CD^2 = AC^2 + BD^2$
b. $AB^2 - BD^2 = AC^2 - CD^2$ Solution
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)
Question 7. In a quadrilateral ABCD, given that $\angle A + \angle D = 90^0$. Prove that $AC^2 + BD^2 = AD^2 + BC^2$ Solution
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)
$AC^2 + BD^2 = AD^2 + BC^2$
Question 8. Prove that a line drawn through the mid- point of one side of a triangle parallel to another side bisects the third side. Question 9. Prove that the line joining the mid – points of any two sides of a triangle is parallel to the third side
Question 10. 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 Solution
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 Question 11. DEF is an equilateral triangle of side $2b$. Find each of its altitudes. Solution
Question 12. Triangle ABC is right- angled at B and D is the mid - point of BC.
Prove that: $AC^2 = 4AD^2 - 3AB^2$ Solution
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)
$AC^2 = 4AD^2 - 3AB^2$
Question 13. 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. Solution
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
Question 14. 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 Solution
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
$\Delta ABC \sim \Delta DEF$