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

Image gives herons formula in simple terms

Herons Forumla in Simple Terms

Image gives herons formula in simple terms

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Art derived from Herons Formula

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 9cm,28cmand35cm (see Fig). Find the cost of polishing the tiles at the rate of 50ppercm2.

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+352cm=36cm

  • Therefore, area of each of the triangular tiles

    • =s(sa)(sb)(sc)

    • =36(3628)(369)(3635)cm2

    • =36×8×27×1cm2

    • =7776

    • =366cm2

    • =88.18cm2

  • Therefore, total area of 16tiles

    • =16×88.2cm2

    • =1411.2cm2

  • Cost of polishing tiles = 50ppercm2

  • Therefore, total cost of polishing tiles

    • =Rs.(1411.2×0.5)

    • =Rs.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,

  • SBQP be the trapezium with parallel sides SB=25mandPQ=10m and the non-parallel sides SP=13mandBQ=14m.

Construction,

  • Draw QASB

  • Draw QR||SP , therefore QPSR is now a parallelogram

In ΔBQR ,

  • BQ=14m,QR=SP=13m and

  • SB=SR+RBBR=SBSR=2510=15m (QPSRisaparallelogramSR=PQ=10cm) .

  • Therefore, semi perimeter of the ΔBQR=BR+QR+BQ2=15+13+142m=21m

Therefore, by Herons formula, area of the ΔBQR

  • =s(sa)(sb)(sc)

  • =21(2114)(2113)(2115)m2

  • =21×7×8×6m2

  • =7056

  • =84m2

Area of the parallelogram SRQP

  • Base×Altitude

  • SR×QA=10×565

  • 112m2

Therefore, area of the trapezium SBQP

  • AreaofSRQP+AreaofΔBQR

  • (112+84)m2

  • 196m2

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