S.chand books class 8 maths solution chapter 24 Circle exercise 24 A

EXERCISE 24 A

---

Q1 | Ex-24A | Class 8 | S.Chand | Composite maths | chapter 24 | Circle | myhelper

OPEN IN YOUTUBE

Question 1

Calculate the length marked by x In each of the following diagrams. the unit of length being cm. (O is the centre in each diagram)

Sol :

 

(i)

In a circle , perpendicular from the center to a chord bisects the chord. 

So, base $\frac{16}{2}=8$

In Δ, Using Pythagoras theorem

10²=8²+x²

x²=100-64

x=√36

x=6cm

(ii)

In a circle , perpendicular from the center to a chord bisects the chord. 

So, base $\frac{6}{2}=3$

In Δ, Using Pythagoras theorem

x²=(1.6)²+3²

x²=2.56+9

x²=11.56

x=3.4cm

(iii)

In a circle , perpendicular from the center to a chord bisects the chord. 

So, base $\frac{x}{2}$

In Δ, Using Pythagoras theorem

(6.5)²=($\frac{x}{2}$)²+(2.5)²

$\left(\frac{65}{10}\right)^2-\left(\frac{25}{10}\right)^2=\frac{x^2}{2^2}$

$\left(\frac{13}{2}\right)^2-\left(\frac{5}{2}\right)^2=\frac{x^2}{4}$

$\frac{169}{4}-\frac{25}{4}=\frac{x^2}{4}$

$\frac{1}{4}\times (169-25)=\frac{x^2}{4}$

(169-25)=x²

√144=x

x=12 cm

---

Q2 | Ex-24A | Class 8 | S.Chand | Composite maths | chapter 24 | Circle | myhelper

OPEN IN YOUTUBE

Question 2

In a circle with centre O and radius 13cm , AB is a chord. If OM perpendicular AB and OM=5cm , what is the length of AB ?

Sol :

Lets find base AM

In ΔAMO , Using Pythagoras theorem

$A O^{2}=A M^{2}+O M^{2}$

$13^{2}=AM^{2}+5^{2}$

$169=A M^{2}+25$

$A M=\sqrt{169-25}=\sqrt{144}$

AM=12

In a circle , ⟂ from the centre to a cord bisects the cord. So, length of cord is 

=2AM

=2×12

=24 cm

---

Q3 | Ex-24A | Class 8 | S.Chand | Composite maths | chapter 24 | Circle | myhelper

OPEN IN YOUTUBE

Question 3

In a circle of radius 17 cm, find the distance of a chord of length 16cm from the centre.

Sol :

In a circle, ⟂ from the center to a cord bisects the cord, so base $=\frac{16}{2}=8$

In ΔOAB, Using Pythagoras theorem

$\begin{aligned} O A^{2} &=AB^{2}+O B^{2} \\ 17^{2} &=8^{2}+O B^{2} \\ 289 &=64+O B^{2} \\ O B &=\sqrt{289-64} \\ &=\sqrt{225}\end{aligned}$

$\begin{array}{l|l}5 & 225 \\\hline 5 & 45 \\\hline 3 & 9 \\\hline 3 & 3 \\ \hline &1\end{array}$

OB=15 cm

---

Q4 | Ex-24A | Class 8 | S.Chand | Composite maths | chapter 24 | Circle | myhelper

OPEN IN YOUTUBE

Question 4

Circle are concentric with centre O. Find AC if OM=8cm , OC=10cm and OB=17cm

Sol :

In ΔOMC, Using Pythagoras theorem

$O C^{2}=O M^{2}+C M^{2}$

$10^{2}=P^{2}+C M^{2}$

$100-64=C M^{2}$

$CM=\sqrt{36}$

CM=6 cm

In ΔOMB, Using Pythagoras Theorem

$\begin{aligned} O B^{2}=& O M^{2}+M B^{2} \\ 17^{2}=& 8^{2}+M B^{2} \\ 289=& 64+M B^{2} \\ M B^{2}=&289-64 \\ M B &=\sqrt{225} \end{aligned}$

MB=15 cm

Also, AM=MB=15

AC=15-6=9 cm

---

Q5 | Ex-24A | Class 8 | S.Chand | Composite maths | chapter 24 | Circle | myhelper

OPEN IN YOUTUBE

Question 5

A chord , distance 2cm from the centre , of a circle is 18cm long. Calculate the length of a chord of the same circle which is 6cm distance from the centre.

Sol :

Let radius of circle be r

AB=18 cm

MB=9 cm

OM=2 cm

OB=r

$r^{2}=9^{2}+2^{2}$

$=\sqrt{81+4}=\sqrt{85}$

Now, ON=6 cm , Let ND=x

$6^{2}+x^{2}=r^{2}$

$6^{2}+x^{2}=85$

$x^{2}=85-36$

$x=\sqrt{49}$

x=7 cm

Length of chord=CD=2x=2×7=14 cm