FORMULA, we have area of TRIANGLE \[=\SQRT{S(S-a)(S-b)(S-c)}\] Here, \[S=\FRAC{11+15+16}{2}=\frac{42}{2}=21\,cm.\] \[\therefore \] Area\[=\sqrt{21(10)(6)(5)}=30\sqrt{7}\,sq.\,cm.\] \[\therefore \] Height\[=\frac{30}{8}\sqrt{7}=\frac{15\sqrt{7}}{4}cm.\]

"> FORMULA, we have area of TRIANGLE \[=\SQRT{S(S-a)(S-b)(S-c)}\] Here, \[S=\FRAC{11+15+16}{2}=\frac{42}{2}=21\,cm.\] \[\therefore \] Area\[=\sqrt{21(10)(6)(5)}=30\sqrt{7}\,sq.\,cm.\] \[\therefore \] Height\[=\frac{30}{8}\sqrt{7}=\frac{15\sqrt{7}}{4}cm.\]

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The sides of a triangle are 11 cm, 15 cm and 16cm. The altitude to largest side is

7th Class Mathematics in 7th Class 11 months ago

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By Heron's FORMULA, we have area of TRIANGLE \[=\SQRT{S(S-a)(S-b)(S-c)}\] Here, \[S=\FRAC{11+15+16}{2}=\frac{42}{2}=21\,cm.\] \[\therefore \] Area\[=\sqrt{21(10)(6)(5)}=30\sqrt{7}\,sq.\,cm.\] \[\therefore \] Height\[=\frac{30}{8}\sqrt{7}=\frac{15\sqrt{7}}{4}cm.\]

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