NCERT Class 9 Solutions: Heron's Formula (Chapter 12) Exercise 12.2 Part 5

Herons Forumla in Simple Terms

Image gives herons formula in simple terms

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Herons Formula Illustration

Art derived from Herons Formula

Q-8 A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 c m , 28 c m a n d 35 c m (see Fig). Find the cost of polishing the tiles at the rate of 50 p p e r c m 2 .

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Quadrilateral poly2 Quadrilateral poly2: Polygon A, E, F, G Quadrilateral poly3 Quadrilateral poly3: Polygon A, H, I, J Quadrilateral poly4 Quadrilateral poly4: Polygon A, K, L, M Quadrilateral poly5 Quadrilateral poly5: Polygon A, N, O, P Quadrilateral poly6 Quadrilateral poly6: Polygon A, Q, R, S Quadrilateral poly7 Quadrilateral poly7: Polygon A, T, U, V Quadrilateral poly8 Quadrilateral poly8: Polygon A, W, Z, A_1 Triangle poly9 Triangle poly9: Polygon U, B_1, T Triangle poly10 Triangle poly10: Polygon S, B_1, R Triangle poly11 Triangle poly11: Polygon K_1, L_1, M_1 Triangle poly12 Triangle poly12: Polygon N_1, O_1, P_1 Triangle poly13 Triangle poly13: Polygon Q_1, R_1, S_1 Triangle poly14 Triangle poly14: Polygon T_1, U_1, V_1 Triangle poly15 Triangle poly15: Polygon V_1, W_1, Z_1 Triangle poly16 Triangle poly16: Polygon A_2, B_2, C_2 Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment a_1 Segment a_1: Segment [A, E] of Quadrilateral poly2 Segment e Segment e: Segment [E, F] of Quadrilateral poly2 Segment f Segment f: Segment [F, G] of Quadrilateral poly2 Segment g Segment g: Segment [G, A] of Quadrilateral poly2 Segment a_2 Segment a_2: Segment [A, H] of Quadrilateral poly3 Segment h Segment h: Segment [H, I] of Quadrilateral poly3 Segment i Segment i: Segment [I, J] of Quadrilateral poly3 Segment j Segment j: Segment [J, A] of Quadrilateral poly3 Segment a_3 Segment a_3: Segment [A, K] of Quadrilateral poly4 Segment k Segment k: Segment [K, L] of Quadrilateral poly4 Segment l Segment l: Segment [L, M] of Quadrilateral poly4 Segment m Segment m: Segment [M, A] of Quadrilateral poly4 Segment a_4 Segment a_4: Segment [A, N] of Quadrilateral poly5 Segment n Segment n: Segment [N, O] of Quadrilateral poly5 Segment o Segment o: Segment [O, P] of Quadrilateral poly5 Segment p Segment p: Segment [P, A] of Quadrilateral poly5 Segment a_5 Segment a_5: Segment [A, Q] of Quadrilateral poly6 Segment q Segment q: Segment [Q, R] of Quadrilateral poly6 Segment r Segment r: Segment [R, S] of Quadrilateral poly6 Segment s Segment s: Segment [S, A] of Quadrilateral poly6 Segment a_6 Segment a_6: Segment [A, T] of Quadrilateral poly7 Segment t Segment t: Segment [T, U] of Quadrilateral poly7 Segment u Segment u: Segment [U, V] of Quadrilateral poly7 Segment v Segment v: Segment [V, A] of Quadrilateral poly7 Segment a_7 Segment a_7: Segment [A, W] of Quadrilateral poly8 Segment w Segment w: Segment [W, Z] of Quadrilateral poly8 Segment z_1 Segment z_1: Segment [Z, A_1] of Quadrilateral poly8 Segment a_8 Segment a_8: Segment [A_1, A] of Quadrilateral poly8 Segment f_1 Segment f_1: Segment [B_1, C_1] Segment g_1 Segment g_1: Segment [B_1, D_1] Segment h_1 Segment h_1: Segment [B_1, E_1] Segment i_1 Segment i_1: Segment [B_1, F_1] Segment j_1 Segment j_1: Segment [B_1, G_1] Segment k_1 Segment k_1: Segment [B_1, H_1] Segment l_1 Segment l_1: Segment [B_1, I_1] Segment m_1 Segment m_1: Segment [B_1, J_1] Segment t_1 Segment t_1: Segment [U, B_1] of Triangle poly9 Segment u_1 Segment u_1: Segment [B_1, T] of Triangle poly9 Segment b_1 Segment b_1: Segment [T, U] of Triangle poly9 Segment r_1 Segment r_1: Segment [S, B_1] of Triangle poly10 Segment s_1 Segment s_1: Segment [B_1, R] of Triangle poly10 Segment b_2 Segment b_2: Segment [R, S] of Triangle poly10 Segment m_2 Segment m_2: Segment [K_1, L_1] of Triangle poly11 Segment k_2 Segment k_2: Segment [L_1, M_1] of Triangle poly11 Segment l_2 Segment l_2: Segment [M_1, K_1] of Triangle poly11 Segment p_1 Segment p_1: Segment [N_1, O_1] of Triangle poly12 Segment n_1 Segment n_1: Segment [O_1, P_1] of Triangle poly12 Segment o_1 Segment o_1: Segment [P_1, N_1] of Triangle poly12 Segment s_2 Segment s_2: Segment [Q_1, R_1] of Triangle poly13 Segment q_1 Segment q_1: Segment [R_1, S_1] of Triangle poly13 Segment r_2 Segment r_2: Segment [S_1, Q_1] of Triangle poly13 Segment v_1 Segment v_1: Segment [T_1, U_1] of Triangle poly14 Segment t_2 Segment t_2: Segment [U_1, V_1] of Triangle poly14 Segment u_2 Segment u_2: Segment [V_1, T_1] of Triangle poly14 Segment z_2 Segment z_2: Segment [V_1, W_1] of Triangle poly15 Segment v_2 Segment v_2: Segment [W_1, Z_1] of Triangle poly15 Segment w_1 Segment w_1: Segment [Z_1, V_1] of Triangle poly15 Segment c_2 Segment c_2: Segment [A_2, B_2] of Triangle poly16 Segment a_9 Segment a_9: Segment [B_2, C_2] of Triangle poly16 Segment b_3 Segment b_3: Segment [C_2, A_2] of Triangle poly16 9 cm text1 = "9 cm" 9 cm text2 = "9 cm" 28 cm text3 = "28 cm" 28 cm text4 = "28 cm" 35 cm text5 = "35 cm" a Floral Design Made of Triangles

A floral design is made with 16 triangular tiles

Solution:

Semi perimeter of each of the triangular shapes = 28 + 9 + 35 2 c m = 36 c m

Therefore, area of each of the triangular tiles

Therefore, total area of 16 t i l e s

Cost of polishing tiles = 50 p p e r c m 2

Therefore, total cost of polishing tiles

= R s . ( 1411.2 × 0.5 )

= R s . 705.6

Q-9 A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. find the area of the field.

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon P, Q, R, S Triangle poly2 Triangle poly2: Polygon Q, B, R Segment a Segment a: Segment [P, Q] of Quadrilateral poly1 Segment b Segment b: Segment [Q, R] of Quadrilateral poly1 Segment c Segment c: Segment [R, S] of Quadrilateral poly1 Segment d Segment d: Segment [S, P] of Quadrilateral poly1 Segment c_1 Segment c_1: Segment [Q, B] of Triangle poly2 Segment b_1 Segment b_1: Segment [B, R] of Triangle poly2 Segment e Segment e: Segment [R, Q] of Triangle poly2 Segment f Segment f: Segment [Q, A] Point P P = (-2.9, 3.98) Point P P = (-2.9, 3.98) Point P P = (-2.9, 3.98) Point Q Q = (1.26, 3.98) Point Q Q = (1.26, 3.98) Point Q Q = (1.26, 3.98) Point R R = (0.06, 1.44) Point R R = (0.06, 1.44) Point R R = (0.06, 1.44) Point S S = (-4.04, 1.44) Point S S = (-4.04, 1.44) Point S S = (-4.04, 1.44) Point B B = (2.66, 1.44) Point B B = (2.66, 1.44) Point B B = (2.66, 1.44) Point A Point A: Point on b_1 Point A Point A: Point on b_1 Point A Point A: Point on b_1 10 m text1 = "10 m" 14 m text2 = "14 m" 25 m text3 = "25 m" 13 m text4 = "13 m" SBQP Is the Required Trapezium

SBQP is the trapezium with parallel sides SB=25mandPQ=10mand the non-parallel sides SP=13mandBQ=14m.

Given,

Construction,

In Δ B Q R ,

B Q = 14 m , Q R = S P = 13 m and

S B = S R + R B ⇒ B R = S B − S R = 25 − 10 = 15 m ( ∵ Q P S R i s a p a r a l l e l o g r a m S R = P Q = 10 c m ) .

Therefore, semi perimeter of the Δ B Q R = B R + Q R + B Q 2 = 15 + 13 + 14 2 m = 21 m

Therefore, by Herons formula, area of the Δ B Q R

Area of the parallelogram S R Q P

B a s e × A l t i t u d e

S R × Q A = 10 × 56 5

112 m 2

Therefore, area of the trapezium S B Q P

A r e a o f S R Q P + A r e a o f Δ B Q R

( 112 + 84 ) m 2

196 m 2