Class 10 Maths: Triangles – Extra Questions with Answers

Ans: B

Solution:
Area = (side² * √3) / 4
Area = (9² * √3) / 4
Area = (81 * √3) / 4
Area = 35.07 cm² (approx.)
Answer: 35.07 cm²

Ans: C

Solution:
(building height) / 30 = 15 / 10
building height = (15/10) * 30
building height = 45

Answer: 45

Ans: C

Solution:
Since $\triangle ABC \sim \triangle DEF$, the ratios of corresponding sides are equal. Thus, $\frac{DE}{AB} = \frac{EF}{BC} = \frac{DF}{AC}$. We have $\frac{DE}{AB} = \frac{3}{6} = \frac{1}{2}$.
The perimeter of $\triangle ABC$ is $6+8+10=24$. Since the sides of $\triangle DEF$ are half the length of the corresponding sides in $\triangle ABC$, the perimeter of $\triangle DEF$ is half the perimeter of $\triangle ABC$.
Perimeter of $\triangle DEF = \frac{1}{2} \times 24 = 12$.

Answer: 12

Ans: D

Solution:
The diagonals of a rhombus bisect each other at right angles. Thus, the diagonals divide the rhombus into four congruent right triangles. The legs of each right triangle are half the lengths of the diagonals, so they are 5 cm and 12 cm. The hypotenuse of each triangle is a side of the rhombus. Using the Pythagorean theorem, the side length is $\sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm. The perimeter of the rhombus is 4 times the side length.
Answer: 52 cm

Ans: D

Solution: The diagonals of a rhombus bisect each other at right angles. Therefore, they divide the rhombus into four congruent right triangles. The legs of each right triangle are half the lengths of the diagonals, which are 3 cm and 4 cm. The hypotenuse (a side of the rhombus) has a length of 5 cm (using the Pythagorean theorem: 3^2 + 4^2 = 9 + 16 = 25, and the square root of 25 is 5). Since all four sides are equal, the sum of the squares of the sides is 4 * 5^2 = 4 * 25 = 100.

Answer: 100

Ans: B

Solution:
Using Pythagorean theorem, $AB = \sqrt{AC^2 + CB^2} = \sqrt{(3x)^2 + (2(x+8))^2} = \sqrt{9x^2 + 4x^2 + 64x + 256} = \sqrt{13x^2 + 64x + 256}$. The direct route AB is 30 km. So, $\sqrt{13x^2 + 64x + 256} = 30$, which gives $13x^2 + 64x + 256 = 900$. Then, $13x^2 + 64x – 644 = 0$. Solving this quadratic equation, we get $x = \frac{-64 \pm \sqrt{64^2 – 4*13*(-644)}}{2*13} = \frac{-64 \pm \sqrt{4096 + 33488}}{26} = \frac{-64 \pm \sqrt{37584}}{26} = \frac{-64 \pm 193.87}{26}$. Since x must be positive, $x = \frac{-64 + 193.87}{26} \approx \frac{129.87}{26} \approx 4.995$. Then, $AC = 3x \approx 3(5) = 15$ km and $CB = 2(x+8) \approx 2(5+8) = 26$ km. The original distance is $AC + CB = 3x + 2(x+8) = 5x + 16 \approx 5(5) + 16 = 41$ km. The saved distance is $41 – 30 = 11$ km. $AC = 3x$. $CB = 2(x+8)$. Then $AC^2 + CB^2 = AB^2$, so $(3x)^2 + (2(x+8))^2 = 30^2$. $9x^2 + 4(x^2 + 16x + 64) = 900$. $9x^2 + 4x^2 + 64x + 256 = 900$. $13x^2 + 64x – 644 = 0$. Then, $(x-3.79)(13x + 169.8) = 0$. Then $x = 3.79$. Thus $AC = 3*3.79 \approx 11.37$. $CB = 2*(3.79 + 8) = 23.58$. Total Distance $11.37 + 23.58 = 34.95$. Distance saved: 34.95 – 30 = 4.95.
$\sqrt{13x^2 + 64x + 256} = 30$
$13x^2 + 64x + 256 = 900$
$13x^2 + 64x – 644 = 0$
$x \approx 3.79$
$AC = 3(3.79) \approx 11.37$
$CB = 2(3.79+8) = 2(11.79) = 23.58$
Total = 34.95
Saved = 34.95 – 30 = 4.95
The correct solution needs to be based on $x$.
Original distance = $3x + 2(x+8) = 5x+16$.
$13x^2 + 64x + 256 = 900$. $13x^2 + 64x -644 = 0$. $x \approx 3.79$.
Original = $5(3.79) + 16 = 18.95+16 = 34.95$. Saved = 34.95-30 = 4.95.

Answer: 4.95

Ans: B

Solution:
Since P and Q are midpoints, $CQ = \frac{1}{2}CB$ and $CP = \frac{1}{2}CA$. By Pythagorean theorem, $AQ^2 = AC^2 + CQ^2 = AC^2 + \frac{1}{4}BC^2$ and $BP^2 = BC^2 + CP^2 = BC^2 + \frac{1}{4}AC^2$. Therefore, $AQ^2 + BP^2 = \frac{5}{4}(AC^2 + BC^2)$. Again, by the Pythagorean theorem, $AC^2 + BC^2 = AB^2 = 100$. Thus $AQ^2 + BP^2 = \frac{5}{4} \cdot 100 = 125$.
Answer: 125

Ans: D

Solution:
Let $\angle YXZ = \theta$.
Using the Law of Cosines on $\triangle XYZ$, we have
$YZ^2 = XY^2 + XZ^2 – 2(XY)(XZ) \cos \theta$.
Since $XY = XZ$, we have $YZ^2 = XY^2 + XY^2 – 2(XY)(XY) \cos \theta = 2XY^2 – 2XY^2 \cos \theta$.
We are given $YZ^2 = 2XY^2$.
Substituting this into the equation gives $2XY^2 = 2XY^2 – 2XY^2 \cos \theta$.
This simplifies to $0 = -2XY^2 \cos \theta$, which implies $\cos \theta = 0$.
Therefore, $\theta = 90^\circ$.

Answer: 90

Ans: D

Solution:
Since the triangles are similar, the ratio of their perimeters equals the ratio of their corresponding sides.
DE/GH = Perimeter of DEF / Perimeter of GHI
DE/9 = 40/30
DE = (40/30) * 9
DE = 12

Answer: 12 cm

Ans: D

Solution:
Let the height of the intersection be $h$. Using similar triangles, we have $\frac{h}{12} + \frac{h}{8} = 1$, where 1 is the ratio of the horizontal distances. Thus, $\frac{2h+3h}{24} = \frac{5h}{24} = 1$, which is incorrect.

Let the flagpoles be at $(0,12)$ and $(15,8)$. The equations of the cables are:
Cable 1: $y = -\frac{12}{15}x + 12 = -\frac{4}{5}x + 12$.
Cable 2: $y = \frac{8}{15}x$.
Intersection: $\frac{8}{15}x = -\frac{4}{5}x + 12 \implies 8x = -12x + 180 \implies 20x = 180 \implies x = 9$.
$y = \frac{8}{15} (9) = \frac{72}{15} = \frac{24}{5} = 4.8$.

Answer: 4.8