Pythagoras Theorem PYQs
Overview
This page provides comprehensive Class 10 Maths Pythagoras Theorem PYQs | Triangles. Pythagoras Theorem previous year questions for Class 10 Maths Triangles. Practice CBSE board PYQs with step-by-step solutions on SJMaths.
CBSE Class 10 PYQs with Step-by-Step Solutions
Qpyth1
2024
00:00
In a right-angled triangle $ABC$, right-angled at $B$, if $AB = 5\text{ cm}$ and $BC = 12\text{ cm}$, find the length of hypotenuse $AC$.
By Pythagoras Theorem in $\triangle ABC$: $AC^2 = AB^2 + BC^2$
$AC^2 = 5^2 + 12^2 = 25 + 144 = 169$
$AC = \sqrt{169} = 13\text{ cm}$
Final Answer: 13 cm (Option a)
Qpyth2
2023
00:00
A ladder $10\text{ m}$ long reaches a window $8\text{ m}$ above the ground. Find the distance of the foot of the ladder from the base of the wall.
Let window height $h = 8\text{ m}$, ladder $L = 10\text{ m}$, distance $= x\text{ m}$.
By Pythagoras Theorem: $x^2 + 8^2 = 10^2 \implies x^2 + 64 = 100 \implies x^2 = 36 \implies x = 6\text{ m}$.
Final Answer: 6 m
Qpyth3
2025
00:00
The perimeter of an isosceles triangle is $32\text{ cm}$. If each equal side is $\left(\frac{5}{6}\right)^{\text{th}}$ of the base, find the area of the triangle.
Step 1: Let base $= b$. Equal sides $a = \frac{5}{6}b$.
Step 2: Perimeter $= b + \frac{5}{6}b + \frac{5}{6}b = \frac{8}{3}b = 32 \implies b = 12\text{ cm}$.
Step 3: Equal sides $a = \frac{5}{6} \times 12 = 10\text{ cm}$.
Step 4: Altitude $h = \sqrt{a^2 - (b/2)^2} = \sqrt{10^2 - 6^2} = \sqrt{64} = 8\text{ cm}$.
Step 5: Area $= \frac{1}{2} \times b \times h = \frac{1}{2} \times 12 \times 8 = 48\text{ cm}^2$.
Final Answer: 48 cm²
Qpyth4
2025
00:00
The perimeter of a right triangle is $60\text{ cm}$ and its hypotenuse is $25\text{ cm}$. Find the lengths of other two sides of the triangle.
Step 1: Let the two perpendicular sides be $x$ and $y$.
Step 2: $x + y + 25 = 60 \implies x + y = 35 \implies y = 35 - x$.
Step 3: By Pythagoras Theorem: $x^2 + y^2 = 25^2 = 625$.
Step 4: $x^2 + (35 - x)^2 = 625 \implies 2x^2 - 70x + 1225 = 625 \implies x^2 - 35x + 300 = 0$.
Step 5: $(x - 15)(x - 20) = 0 \implies x = 15\text{ cm}$ or $x = 20\text{ cm}$.
Thus, the other two sides are $15\text{ cm}$ and $20\text{ cm}$.
Final Answer: 15 cm and 20 cm
Qpyth5
2025
00:00
In $\triangle ABC$, if $AD \perp BC$ and $AD^2 = BD \times DC$, then prove that $\angle BAC = 90^\circ$.
Step 1: In right $\triangle ADB$: $AB^2 = AD^2 + BD^2$.
Step 2: In right $\triangle ADC$: $AC^2 = AD^2 + CD^2$.
Step 3: Adding both equations: $AB^2 + AC^2 = 2AD^2 + BD^2 + CD^2$.
Step 4: Substitute $AD^2 = BD \cdot CD$:
$$AB^2 + AC^2 = 2(BD \cdot CD) + BD^2 + CD^2 = (BD + CD)^2 = BC^2$$
Step 5: By converse of Pythagoras Theorem, $\angle BAC = 90^\circ$. (Proved)
Final Answer: ∠BAC = 90° (Proved)
Qpyth6
2023
00:00
In a $\triangle PQR$, $N$ is a point on $PR$, such that $QN \perp PR$. If $PN \times NR = QN^2$, prove that $\angle PQR = 90^\circ$.
Step 1: In right $\triangle PNQ$: $PQ^2 = PN^2 + QN^2$.
Step 2: In right $\triangle RNQ$: $QR^2 = NR^2 + QN^2$.
Step 3: Adding both: $PQ^2 + QR^2 = PN^2 + NR^2 + 2 QN^2$.
Step 4: Given $QN^2 = PN \cdot NR \implies PQ^2 + QR^2 = PN^2 + NR^2 + 2(PN \cdot NR) = (PN + NR)^2 = PR^2$.
Step 5: By converse of Pythagoras Theorem, $\angle PQR = 90^\circ$. (Proved)
Final Answer: ∠PQR = 90° (Proved)