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Let the side of the smaller square be x cm and side of the larger square be y cm. Then, According to the given conditions, we have y 2 – 2x 2 = 14 ...(i) and 2y 2 + 3x 2 = 203 ...(ii) Substituting y 2 = 14 + 2x 2 From (i) in (ii), we get 2(14 + 2x 2 ) + 3x 2 = 203 ⇒ 28 + 4x 2 + 3x 2 = 203 ⇒ 7x 2 = 175 ⇒ x 2 = 25 ⇒ x = ±5 ∴ x = 5 [x = –5 is rejected] From (i), y 2 –2x 2 = 14 ⇒ y 2 – 2(5) 2 = 14 ⇒ y 2 – 50 = 14 ⇒ y 2 = 64 ⇒ y = ±8 ⇒ y = 8 [y = –8 is rejected] Hence, side of smaller square = 5 cm and side of larger square = 8 cm.

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Quadratic Equations

Long Answer Type

191. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm. Find the other two sides.

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193. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 sq cm. However, if twice the area of larger square is added to three times the area of the smaller square, the result is 203 sq. cm. Find the sides of the two squares.

Let the side of the smaller square be x cm and side of the larger square be y cm. Then,
According to the given conditions, we have
y² – 2x² = 14 ...(i)
and 2y² + 3x² = 203 ...(ii)
Substituting y² = 14 + 2x²From (i) in (ii), we get
2(14 + 2x²) + 3x²= 203
⇒ 28 + 4x² + 3x² = 203
⇒ 7x² = 175
⇒ x^{2 =} 25
⇒ x = ±5
∴ x = 5 [x = –5 is rejected]
From (i), y² –2x² = 14
⇒ y²– 2(5)² = 14
⇒ y² – 50 = 14
⇒ y² = 64
⇒ y = ±8
⇒ y = 8 [y = –8 is rejected]
Hence, side of smaller square = 5 cm
and side of larger square = 8 cm.

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Short Answer Type

194. The hypotenuse of a right triangle is

3 square root of 5 space cm.

If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be 15 cm. Find the length of each side.

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Long Answer Type

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197. Two pipes running together can fill a cistern in

space space 3 1 over 13 space minutes.

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