CBSE Class 10 Mathematics Triangles Assignment Set 16

Exercise-1 Subjective: Areas of Similar Triangles

Question 1. If the areas of two similar triangles are in the ratio 25 : 64, write the ratio of their corresponding sides.
Answer:
Ratio of sides \( = \sqrt{\dfrac{25}{64}} = \) 5 : 8.
In simple words: Take the square root of the area ratio.

Teacher's Note:
1. The area ratio is the square of the side ratio.

Teacher's Note:
1. Triangle ADE is similar to triangle ABC.
2. Triangle DEF is similar to triangle CBF with the same ratio.

Question 3. D, E and F are the mid-points of the sides AB, BC and CA of \( \triangle ABC \). Find \( \dfrac{ar(\triangle DEF)}{ar(\triangle ABC)} \).
Answer:
By the mid-point theorem \( DE = \dfrac{1}{2}AC \), \( EF = \dfrac{1}{2}AB \), \( DF = \dfrac{1}{2}BC \), so \( \triangle DEF \sim \triangle CAB \) (SSS) with ratio \( \dfrac{1}{2} \).
\( \dfrac{ar(DEF)}{ar(ABC)} = \left(\dfrac{1}{2}\right)^2 = \) \( \dfrac{1}{4} \).
In simple words: The medial triangle has half the sides, so a quarter of the area.

Teacher's Note:
1. The segment joining mid-points is half the third side.

Question 4. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. Using this, do the following: the diagonals of a trapezium ABCD, with \( AB \parallel DC \), intersect each other at the point O. If AB = 2CD, find the ratio of the area of \( \triangle AOB \) to the area of \( \triangle COD \).
Answer:
Theorem. Let \( \triangle ABC \sim \triangle PQR \); draw \( AM \perp BC \) and \( PN \perp QR \). \( \triangle ABM \sim \triangle PQN \) (AA) gives \( \dfrac{AM}{PN} = \dfrac{AB}{PQ} = \dfrac{BC}{QR} \).
\( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{\frac{1}{2}BC \cdot AM}{\frac{1}{2}QR \cdot PN} = \dfrac{BC}{QR} \cdot \dfrac{BC}{QR} = \dfrac{BC^2}{QR^2} = \dfrac{AB^2}{PQ^2} \).
Trapezium. \( \triangle AOB \sim \triangle COD \) (alternate angles), so \( \dfrac{ar(AOB)}{ar(COD)} = \left(\dfrac{AB}{CD}\right)^2 = 2^2 = \) 4 : 1.
In simple words: Area ratio is the square of the side ratio.

Teacher's Note:
1. Draw the two altitudes to prove the theorem.
2. The triangles at the crossing of the diagonals are similar.

Question 5. AD is an altitude of an equilateral triangle ABC. On AD as base, another equilateral triangle ADE is constructed. Prove that area (\( \triangle ADE \)) : area (\( \triangle ABC \)) = 3 : 4.
Answer:
Let the side of \( \triangle ABC \) be a. Then \( AD = \dfrac{\sqrt{3}}{2}a \).
Both triangles are equilateral, so they are similar: \( \dfrac{ar(ADE)}{ar(ABC)} = \dfrac{AD^2}{AB^2} = \dfrac{\frac{3}{4}a^2}{a^2} = \dfrac{3}{4} \). So the ratio is 3 : 4.
In simple words: Any two equilateral triangles are similar.

Teacher's Note:
1. The altitude of an equilateral triangle is (√3/2) times its side.
2. Area ratio is the square of the side ratio.

Exercise-1 Subjective: Pythagoras Theorem

Teacher's Note:
1. Eliminate AD².

Teacher's Note:
1. The two small triangles are similar to each other.

Question 8. The perpendicular AD on the base BC of \( \triangle ABC \) meets BC at D so that \( 2DB = 3CD \). Prove that \( 5AB^2 = 5AC^2 + BC^2 \).
Answer:
Let DB = 3k and CD = 2k, so BC = 5k. \( AB^2 = AD^2 + 9k^2 \) and \( AC^2 = AD^2 + 4k^2 \).
\( AB^2 - AC^2 = 5k^2 \Rightarrow 5AB^2 - 5AC^2 = 25k^2 = BC^2 \). Hence \( 5AB^2 = 5AC^2 + BC^2 \).
In simple words: Subtract the two Pythagoras relations.

Teacher's Note:
1. Take DB = 3k and CD = 2k.
2. AB² − AC² = DB² − CD².

Question 9. D and E are points on the sides CA and CB respectively of \( \triangle ABC \) right-angled at C. Prove that \( AE^2 + BD^2 = AB^2 + DE^2 \).
Answer:
\( AE^2 = AC^2 + CE^2 \) and \( BD^2 = BC^2 + CD^2 \) (right angle at C).
Adding: \( AE^2 + BD^2 = (AC^2 + BC^2) + (CE^2 + CD^2) = AB^2 + DE^2 \) (using the right triangles ABC and DCE).
In simple words: Add the Pythagoras relations of the four right triangles.

Teacher's Note:
1. The right angle at C is common to four triangles.

Teacher's Note:
1. S is the mid-point of QR.
2. Use the theorem in triangles PQS and PQR.

Question 11. In \( \triangle ABC \), \( \angle ABC = 135^\circ \). Prove that \( AC^2 = AB^2 + BC^2 + 4\,ar(\triangle ABC) \).
Answer:
Draw \( AD \perp CB \) produced, meeting it at D. \( \angle ABD = 45^\circ \), so AD = BD and \( AB^2 = 2AD^2 \).
\( AC^2 = AD^2 + CD^2 = AD^2 + (BC + BD)^2 = (AD^2 + BD^2) + BC^2 + 2BC \cdot BD = AB^2 + BC^2 + 2BC \cdot BD \).
\( ar(ABC) = \dfrac{1}{2} \times BC \times AD = \dfrac{1}{2}BC \cdot BD \), so \( 2BC \cdot BD = 4\,ar(ABC) \). Hence the result.
In simple words: The altitude outside the triangle makes a 45° right triangle.

Teacher's Note:
1. Extend CB to meet the perpendicular from A.
2. AD = BD because the angle at B is 45°.

Teacher's Note:
1. First find the missing side PR.
2. 6, 8, 10 is a right triangle.

Question 13. In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Prove it. Using this, do the following: in an isosceles triangle PQR, PQ = QR and \( PR^2 = 2PQ^2 \). Prove that \( \angle Q \) is a right angle.
Answer:
Theorem. As proved in the previous question: if \( AC^2 = AB^2 + BC^2 \), a right triangle PQR with legs AB and BC has hypotenuse PR = AC, so \( \triangle ABC \cong \triangle PQR \) and \( \angle B = 90^\circ \).
Result. PQ = QR, so \( PQ^2 + QR^2 = 2PQ^2 = PR^2 \). By the converse of the Pythagoras theorem, \( \angle Q = 90^\circ \).
In simple words: Two equal sides squared and added give the square of the third.

Teacher's Note:
1. PQ² + QR² = 2PQ².
2. The side opposite the right angle is PR.