Match the two columns-Column-I -Column-IIA The radius of - a - circle - is - 8 - cm - and- p - 23 - cmthe length of - one of its chords - is 12 - cm - ll - The distance of - the - chord - form the - centre is - 11B Two parallel chords of length 30 - cm - 8 q - 5-196 cm - and - 16 - cm - are drawn on - the opposite sides - of - the centre of - a - circle of radius 17 - cm - IIC The length of - a - chords - which - is - at - a - distance of - r - 5-291 - cm l- 4 - cm - from the centre - of - the circle of - radius 6 - cm - is 11D - An equilateral triangle of - side - 9 - cm - is inscribed 8 s - 8-94 - cm - Choose - the correct - option- e n d a r r a y A- A-r- B-p- C-q- D-sB- A-r- B-s- C-p- D-qC- A-p- B-q- C-r- D-sD- A-r- B-p- C-s- D-q

The correct option is D A r; B p; C s; D q (b) (A)→ r;(B)→ p;(C)→ s;(D)→ q (A) Let PQ be the chord of a circle with centre O and radius 8 cm such that PQ = 12 ...

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Byju's Answer

Standard X

Mathematics

Relation between Inradius and Perimeter of Triangle

Match the two...

Question

Match the two columns:
$\begin{matrix} C o l u m n - I & C o l u m n - I I (A) T h e r a d i u s o f a c i r c l e i s 8 c m a n d & (p) 23 c m t h e l e n g t h o f o n e o f i t s c h o r d s i s 12 c m . T h e d i s t a n c e o f t h e c h o r d f o r m t h e c e n t r e i s (B) T w o p a r a l l e l c h o r d s o f l e n g t h 30 c m & (q) 5.196 c m a n d 16 c m a r e d r a w n o n t h e o p p o s i t e s i d e s o f t h e c e n t r e o f a c i r c l e o f r a d i u s 17 c m . T h e d i s t a n c e b e t w e e n t h e c h o r d s i s (C) T h e l e n g t h o f a c h o r d s w h i c h i s a t a d i s t a n c e o f & (r) 5.291 c m 4 c m f r o m t h e c e n t r e o f t h e c i r c l e o f r a d i u s 6 c m i s (D) A n e q u i l a t e r a l t r i a n g l e o f s i d e 9 c m i s i n s c r i b e d & (s) 8.94 c m i n a c i r c l e . T h e r a d i u s o f t h e c i r c l e i s C h o o s e t h e c o r r e c t o p t i o n : \end{matrix}$

A-r; B-p; C-q; D-s

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A-r; B-p; C-s; D-q

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A-r; B-s; C-p; D-q

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A-p; B-q; C-r; D-s

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Solution

The correct option is B
A-r; B-p; C-s; D-q

(b) $(A) \to r; (B) \to p; (C) \to s; (D) \to q$

(A) Let PQ be the chord of a circle with centre O and radius 8 cm such that PQ = 12 cm. From O, draw $O L ⊥ P Q$ . Join OP. Since the perpendicular from the centre of a circle to a chord bisects the chord. $P L = L Q = \frac{1}{2} P Q = 6 c m$ In right triangle OLP, we have

$O P^{2} = O L^{2} + P L^{2}$
$\Rightarrow 8^{2} = O L^{2} + 6^{2} \Rightarrow O L^{2} = 8^{2} - 6^{2} \Rightarrow O L^{2} = 8^{2} - 6^{2} = 64 - 36 = 28 \Rightarrow O L = \sqrt{28} = 5.291 c m$
Hence, the distance of the chord from the centre is 5.291 cm.

(B) We know that the perpendicular from the centre of a circle to a chord bisects the chord.

In $Δ$ PLO, PL = 15cm and PO = 17cm

By pythagoras theoram,

$P O^{2}$ = $P L^{2}$ + $L O^{2}$

$L O^{2}$ = $15^{2}$ - $17^{2}$

$L O^{2}$ = $225$ - $289$

$L O$ = $8$ cm

Similarly MO = $15$ cm

Therefore, LM = 15 + 8 = 23cm
(C) Let PQ be a chord of a circle with centre O and radius 6 cm. Draw $O L ⊥ P Q$ . Join OP then OL = 4 cm and OP = 6 cm.
From the right-angled ΔOPL, we have
$O P^{2} = O L^{2} + P L^{2} \Rightarrow P L^{2} = O P^{2} - O L^{2} \Rightarrow P L^{2} = 6^{2} - 4^{2} = 36 - 16 = 20 \Rightarrow P L^{2} = 4.472 c m$

Since the perpendicular from the centre of the circle to a chord bisects the chord, we have
$P Q = 2 \times P L = 2 \times 4.472 = 8.94 c m$

(D) Let $Δ$ PQR be an equilateral triangle of side 9 cm. Let PS be one of its medians. Then $P S ⊥ Q R$ and QS = 4.5 cm
$P S = \sqrt{P Q^{2} - Q S^{2}}$
$= \sqrt{9^{2} - {(\frac{9}{2})}^{2} c m} = \frac{9 \sqrt{3}}{2} c m$

In an equilateral triangle, the centroid and circumcenter coincide and PG:PS = 2 : 3
$∴ R a d i u s = P G = \frac{2}{3} P S = \frac{2}{3} \times \frac{9 \sqrt{3}}{2}$
$= 3 \sqrt{3} = 5.196 c m$

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