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1.	What is the area of the triangle having sides equal to 10 cm, 16 cm and 24 cm?(a) \(25\sqrt{15} cm^2\)(b) \(35\sqrt{17} cm^2\)(c) \(15\sqrt{13} cm^2\)(d) \(15\sqrt{15} cm^2\)I got this question in an international level competition.Enquiry is from Heron’s Formula & Area of a Triangle by Heron’s Formula topic in portion Heron’s Formula of Mathematics – Class 9
Answer» The CORRECT OPTION is (d) \(15\sqrt{15} cm^2\) Easy explanation: a = 10, b = 16 and c = 24 s = \(\frac{a+b+c}{2}=\frac{10+16+24}{2} = \frac{50}{2}\) = 25 According to heron’s formula, area of the TRIANGLE = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{25(25-10)(25-16)(25-24)}\) = \(\sqrt{25159*1}\) = \(15\sqrt{15} cm^2\)

2.	Which of the following formula is used for finding the area of triangle?(a) base * height(b) \(\frac{1}{2}\) * base * height(c) 2 * base * height(d) base^2 * heightI had been asked this question by my school principal while I was bunking the class.Enquiry is from Heron’s Formula & Area of a Triangle by Heron’s Formula in section Heron’s Formula of Mathematics – Class 9
Answer» The correct OPTION is (b) \(\frac{1}{2}\) * base * height For explanation: As SHOWN in the DIAGRAM below, area of the triangle is given by \(\frac{1}{2}\) * base * height where, base = BC and height is AD.

3.	A triangle and a parallelogram has same base and same area as shown in the diagram below. Dimensions of triangle are 28cm, 26cm and 30cm with 28cm being the base. What is the height of the parallelogram?(a) 15cm(b) 10cm(c) 12cm(d) 18cmI have been asked this question during an interview.The doubt is from Application of Heron’s Formula in finding Areas of Quadrilaterals topic in division Heron’s Formula of Mathematics – Class 9
Answer» Right option is (c) 12cm To explain: a = 28, b = 26 and c = 30 cm. s = \(\frac{a+b+c}{2}=\frac{28+26+30}{2} = \frac{84}{2}\) = 42 According to HERON’s formula, area of the TRIANGLE = \(\SQRT{s(s-a)(s-b)(s-c)}\) = \(\sqrt{42(42-28)(42-26)(42-30)}\) = \(\sqrt{42141612}\) = 336 cm^2 It is GIVEN that area of the triangle and the parallelogram is same. Now, we know that area of a parallelogram = BASE perpendicular (height of a parallelogram) Therefore, 28 * height= 336 height = \(\frac{336}{28}\) height = 12cm.

4.	A triangular board having sides 45m, 30m and 35m is used for advertising. One company uses this board for its advertisement for 4 months. How much rent will the company has to pay if the rent is Rs 3500 per m^2?(a) Rs 611800(b) Rs 1835400(c) Rs 611900(d) Rs 1835500The question was posed to me in an interview for internship.The question is from Heron’s Formula & Area of a Triangle by Heron’s Formula topic in section Heron’s Formula of Mathematics – Class 9
Answer» Correct choice is (a) Rs 611800 The explanation: a = 45, B = 30 and C = 35 We know that s = \(\frac{a+b+c}{2} = \frac{45+30+35}{2} = \frac{110}{2}\) = 55 According to heron’s formula, area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{55(55-45)(55-30)(55-35)}\) = \(\sqrt{55102520}\) = 524.40 m^2 Now the rent for 1 m^2 = Rs 3500, then rent for 992.15 m^2 = 524.40 3500 = Rs 18,35,400 The rent which we got above is for one year. The company used the board for 4 MONTHS. Therefore rent for 4 months = Rs \(\frac{18,35,400*4}{12}\) = Rs 6,11,800.

5.	In heron’s formula\(\sqrt{s(s-a)(s-b)*(s-c)}\), what is the value of s if a, b and c are sides of the triangle?(a) \(\frac{a+b+c}{4}\)(b) a+b+c(c) \(\frac{a+b+c}{2}\)(d) 2a+2b+2cI have been asked this question in a job interview.My question is based upon Heron’s Formula & Area of a Triangle by Heron’s Formula in chapter Heron’s Formula of Mathematics – Class 9
Answer» CORRECT OPTION is (C) \(\frac{a+b+c}{2}\) Best EXPLANATION: In heron’s formula\(\sqrt{s(s-a)(s-b)*(s-c)}\), s is the half perimeter of the triangle. Perimeter of the triangle having SIDES a, b and c is a + b + c. Hence, s = half perimeter of the triangle = \(\frac{a+b+c}{2}\).

6.	The sides of a triangle are in the proportion of 2 : 3 : 5 and its perimeter is 200 cm. The area of this triangle is __________ cm^2.(a) 375 \(\sqrt{23}\)(b) 375 \(\sqrt{21}\)(c) 345 \(\sqrt{23}\)(d) 345 \(\sqrt{21}\)This question was posed to me in an online interview.Asked question is from Heron’s Formula & Area of a Triangle by Heron’s Formula topic in section Heron’s Formula of Mathematics – Class 9
Answer» Correct choice is (a) 375 \(\sqrt{23}\) EXPLANATION: Let the side of a triangles be a = 2x, b = 3x and c = 5X. Perimeter of the triangle is 200 cm. Therefore a + b + c = 2x + 3x + 5x = 200 10x = 200 Hence, x = 20 a = 2x = 2(20) = 40 cm b = 3x = 3(30) = 90 cm c = 5x = 5(20) = 100 cm Now, s = \(\frac{a+b+c}{2}=\frac{40+90+100}{2} = \frac{230}{2}\) = 115 According to heron’s formula, area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{115(115-40)(115-90)(115-100)}\) = \(\sqrt{1157525*15}\) = \(375 \sqrt{23}\)

7.	An umbrella is made by stitching 8 triangular pieces of cloth of two different colours, each piece measures 60cm, 60cm and 20cm. How much cloth of each colour is required for the umbrella?(a) \(50\sqrt{35} cm^2\)(b) \(25\sqrt{65} cm^2\)(c) \(50\sqrt{45} cm^2\)(d) \(25\sqrt{55} cm^2\)I have been asked this question in semester exam.Question is from Application of Heron’s Formula in finding Areas of Quadrilaterals in chapter Heron’s Formula of Mathematics – Class 9
Answer» CORRECT option is (a) \(50\sqrt{35} cm^2\) The explanation: s = \(\frac{a+b+c}{2}=\frac{60+60+20}{2} = \frac{140}{2}\) = 70 According to heron’s FORMULA, AREA of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) = \(\sqrt{70(70-60)(70-60)(70-20)}\) = \(\sqrt{701010*50}\) = 100\(\sqrt{35} cm^2\) This area is the combined area of both colours. Hence, area of each colour = \(\frac{100\sqrt{35}}{2}\) = \(50\sqrt{35} cm^2\).