Question

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Answer 1

Answer:

Q1.

Consider Δ QBC. Since it is a right angled triangle, we can use the pythagorean theorem to find the measure of the unknown third side.

Applying the Pythagoras Theorem we get,

⇒ QB² + BC² = QC²

Substituting the known values, we get:

⇒ 12² + BC² = 15²

⇒ 144 + BC² = 225

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = √81 = 9 m.

Hence the measure of side BC is 9 m.

Consider ΔPAC. Applying Pythagoras Theorem, we get:

⇒ PA² + AC² = PC²

⇒ 9² + AC² = 15²

⇒ 81 + AC² = 225

⇒ AC² = 225 - 81

⇒ AC² = 144

⇒ AC = √144 = 12 m

Hence the measure of side AC is 12 m.

Hence the width of road AB = AC + BC

⇒ Width (AB) = 12 + 9 = 21 m.

Q2.

Since it is a right angled triangle, we can apply the Pythagoras Theorem. Hence we get:

⇒ AC² + BC² = AB²

⇒ (5√3)² + 5² = AB²

⇒ (25 × 3) + 25 = AB²

⇒ 75 + 25 = AB²

⇒ 100 = AB²

⇒ AB = √100 = 10 m.

Hence the length of the ladder is 10 m.

Q3.

Since Angle Q is 90 degrees, ΔPQR is a right angled triangle. It is given that the ratio of PQ/PR = 1/2. Cross multiplying the terms we get:

⇒ PR = 2 PQ

Now applying Pythagoras Theorem we get:

⇒ PQ² + QR² = PR²

⇒ PQ² + QR² = ( 2 PQ )²

⇒ PQ² + QR² = 4 PQ²

⇒ QR² = 4 PQ² - PQ²

⇒ QR² = 3 PQ²

⇒ QR = √ (3 PQ²)

⇒ QR = PQ√3

Q4.

For two similar triangles ΔABC and ΔDEF, the relation between areas and their corresponding sides is given as:

$\implies \dfrac{ar(\Delta_1)}{ar(\Delta_2)} = \dfrac{AB^2}{DE^2} = \dfrac{BC^2}{EF^2} = \dfrac{AC^2}{DF^2}$

According to the question,

Ratio of Areas = 25/9.

Ratio of Sides = ?

$\implies \dfrac{25}{9} = \dfrac{AB^2}{PQ^2}\\\\\\\implies \sqrt{\dfrac{25}{9}} = \dfrac{AB}{PQ}\\\\\\\implies \dfrac{AB}{PQ} = \dfrac{5}{3} \:\:(or)\:\:5:3$

Hence the required answer is 5/3.

Q5.

Area 1 (ΔABC) = 144 cm²

Area 2 (ΔPQR) = 81 cm²

Side 1 (BC) = ?

Side 2 (QR) = 27 cm

Using the Area and Sides of similar triangles relation, we get:

⇒ 144/81 = (BC/27)²

Taking Square root on both sides we get:

⇒ √(144/81) = BC/27

⇒ 12/9 = BC/27

⇒ BC = (4/3) × 27

⇒ BC = 4 × 9 = 36 cm.

Hence the length of BC is 36 cm.

Q6.

(DE/PQ) = 4/9

Area(ΔDEF) / Area(ΔPQR) = ?

Using the Area and Sides of similar triangles relation, we get:

⇒ Area(ΔDEF) / Area(ΔPQR) = (DE/PQ)²

⇒ Area(ΔDEF) / Area(ΔPQR) = (4/9)²

⇒ Area(ΔDEF) / Area(ΔPQR) = 16/81

Hence the required ratio is 16 : 81 (or) 16/81.

Q7 and Q8 are answered in the attachments.

Answer 2

Answer:

Consider Δ QBC. Since it is a right angled triangle, we can use the pythagorean theorem to find the measure of the unknown third side.

Applying the Pythagoras Theorem we get,

⇒ QB² + BC² = QC²

Substituting the known values, we get:

⇒ 12² + BC² = 15²

⇒ 144 + BC² = 225

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = √81 = 9 m.

Hence the measure of side BC is 9 m.

Consider ΔPAC. Applying Pythagoras Theorem, we get:

⇒ PA² + AC² = PC²

⇒ 9² + AC² = 15²

⇒ 81 + AC² = 225

⇒ AC² = 225 - 81

⇒ AC² = 144

⇒ AC = √144 = 12 m

Hence the measure of side AC is 12 m.

Hence the width of road AB = AC + BC

⇒ Width (AB) = 12 + 9 = 21 m.

Q2.

Since it is a right angled triangle, we can apply the Pythagoras Theorem. Hence we get:

⇒ AC² + BC² = AB²

⇒ (5√3)² + 5² = AB²

⇒ (25 × 3) + 25 = AB²

⇒ 75 + 25 = AB²

⇒ 100 = AB²

⇒ AB = √100 = 10 m.

Hence the length of the ladder is 10 m.

Q3.

Since Angle Q is 90 degrees, ΔPQR is a right angled triangle. It is given that the ratio of PQ/PR = 1/2. Cross multiplying the terms we get:

⇒ PR = 2 PQ

Now applying Pythagoras Theorem we get:

⇒ PQ² + QR² = PR²

⇒ PQ² + QR² = ( 2 PQ )²

⇒ PQ² + QR² = 4 PQ²

⇒ QR² = 4 PQ² - PQ²

⇒ QR² = 3 PQ²

⇒ QR = √ (3 PQ²)

⇒ QR = PQ√3

Q4.

For two similar triangles ΔABC and ΔDEF, the relation between areas and their corresponding sides is given as:

\implies \dfrac{ar(\Delta_1)}{ar(\Delta_2)} = \dfrac{AB^2}{DE^2} = \dfrac{BC^2}{EF^2} = \dfrac{AC^2}{DF^2}⟹ar(Δ2)ar(Δ1)=DE2AB2=EF2BC2=DF2AC2

According to the question,

Ratio of Areas = 25/9.

Ratio of Sides = ?

\begin{gathered}\implies \dfrac{25}{9} = \dfrac{AB^2}{PQ^2}\\\\\\\implies \sqrt{\dfrac{25}{9}} = \dfrac{AB}{PQ}\\\\\\\implies \dfrac{AB}{PQ} = \dfrac{5}{3} \:\:(or)\:\:5:3\end{gathered}⟹925=PQ2AB2⟹925=PQAB⟹PQAB=35(or)5:3