Answer: Naman pulls in the string at the rate of 5 cm per second. Hence, after 12 seconds the length of the string he will pulled is given by: 12 × 5 = 60 cm or 0.6 m In ∆BMC By...

        Answer: Adding $\Rightarrow A D^{2}+B C \cdot D L+\left(\frac{B C}{2}\right)^{2}$ and $A B^{2}=A D^{2}-B C-D L+\left(\frac{B C}{2}\right)^{2}$, $\begin{aligned} AC^{2}+A...

        Answers: (i) In right triangle ALD, Using Pythagoras theorem, $A C^{2}=A L^{2}+L C^{2}$ $\Rightarrow A D^{2}-D L^{2}+(D L+D C)^{2}$ $\Rightarrow A D^{2}-D L^{2}+\left(D...

          Answer: Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr Distance...

        Answer: ABC as an isosceles triangle, right angled at B. AB = BC Applying Pythagoras theorem in right-angled triangle ABC, ????????2 = ????????2 + ????????2 = 2????????2...

      Answer: Draw AE⊥BC, meeting BC at D. Applying Pythagoras theorem in right-angled triangle AED, ABC is an isosceles triangle and AE is the altitude and we know that the altitude...

      Answers: (i) Adding $b^{2}=p^{2}+a x+\frac{a^{2}}{x}$ and $c^{2}=p^{2}-a x+\frac{a^{2}}{x}$, $b^{2}+c^{2}=p^{2}+a x+\frac{a^{2}}{4}+p^{2}-a x+\frac{a^{2}}{4}$ $b^{2}+c^{2}=2...

      Answers: (i) In right-angled triangle AEC, Applying Pythagoras theorem, ????????2 = ????????2 + ????C2 $b^{2}=h^{2}+\left(x+\frac{a}{2}\right)^{2}$ …(1) In right – angled...

      Answer: Given, ∠???????????? = 90o ???????? ⊥ ????B In ∆ ACB and ∆ CDB ∠???????????? = ∠???????????? = 90o ∠???????????? = ∠???????????? By AA similarity-criterion, ∆ ACB ~ ∆CDB...

      Answer: In right-angled triangle AED, applying Pythagoras theorem, ????????2 = ????????2 + ????????2 … (????) In right-angled triangle AED, applying Pythagoras theorem,...

      Answer: Let ABCD be the rhombus with diagonals meeting at O. AC = 24 cm BD = 10 cm We know that the diagonals of a rhombus bisect each other at angles. Applying Pythagoras...

      Answer: Let ABCD be the rectangle with diagonals AC and BD meeting at O. AB = CD = 30 cm BC = AD = 16 cm Applying Pythagoras theorem in right-angled triangle ABC, we get:...

      Answer: Let ABC be the equilateral triangle with AD as an altitude from A meeting BC at D. D will be the midpoint of BC. Applying Pythagoras theorem in right-angled triangle...

      Answer: Let AD, BE and CF be the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, D, E and F are the midpoint of BC, AC and AB, In right-angled ∆ABD, AB = 2a BD = a...

          Answer: In isosceles ∆ ABC, AB = AC = 2a units BC = a unit Let AD be the altitude drawn from A that meets BC at D. D is the midpoint of BC. BD = BC = ????2...

            Answer: Given, ∆ ABC is an isosceles triangle. AB = AC = 13 cm The altitude from A on BC meets BC at D. D is the midpoint of BC. AD = 5 cm ∆ ????????????...

      Answer: Applying Pythagoras theorem in right-angled triangle POR, ????????2 = ????????2 + ????????2 ????????2 = 62 + 82 =>36 + 64 =>100 ???????? = √100 =>10 ???????? In...

        Answer: Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A. In right triangle ABC By using Pythagoras theorem,...

      Answer: Let the two poles be DE and AB and the distance between their bases be BE. DE = 9 m AB = 14 m BE = 12 m Draw a line parallel to BE from D, meeting AB at C. DC = 12 m AC...

      Answer: Let the height of the window from the ground and the distance of the foot of the ladder from the wall be AB and BC, AB = 20 m BC = 15 m Applying Pythagoras theorem in...

        Answer: Let AB and AC be the ladder and height of the building. Given, AB = 13 m AC = 12 m In right-angled triangle ABC, $A B^{2}=A C^{2}+B C^{2}$ $B...

        Answer: Let the man starts from point D and goes 10 m due south at E. He then goes 24 m due west at F. In right ∆DEF, DE = 10 m EF = 24 m $D F^{2}=E F^{2}+D E^{2}$ $D...

          Answer: Let the man starts from point A and goes 80 m due east to B. From B, he goes 150 m due north to c. In right- angled triangle ABC, $A C^{2}=A B^{2}+B C^{2}$...

Answer: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. P = (a-1) cm, q = 2 √???? ???????? ???????????? ???? = (???? + 1)...

Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 1.4 cm b= 4.8 cm c= 5 cm ????2 + ????2 = (1.4) 2 + (4.8)...

Answers: Given, The sum of the two sides must be equal to the square of the third side. The three sides of the triangle - a, b and c. (i) a = 9 cm b = 16 cm c = 18 cm ????2 + ????2 = 92 + 162...