<\/span><\/h4>\nMark the correct alternative in each of the following:<\/em>
The sides of a triangle are 16 cm, 30 cm, 34 cm. Its area is
(a) 225 cm2<\/sup>
(b) 240 cm2<\/sup>
(c) 225<\/mn>2<\/mn><\/msqrt><\/math>“>225<\/span>2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span>2252<\/span><\/span> cm2<\/sup>
(d) 450 cm2<\/sup><\/p>\n<\/div>\n\n
<\/span> <\/span><\/h4>\n<\/span>ANSWER:<\/span><\/h4>\nThe area of a triangle having sides a<\/em>, b<\/em>, c <\/em>and s<\/em> as semi-perimeter is given by,<\/span><\/span><\/p>\n![\"\"](\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/iit_pretests\/1\/9\/605\/3405\/\/Ex.12.MCQ_Final%20File_html_73551a9a.gif\")
, where<\/span><\/span><\/p>\n![\"\"](\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/iit_pretests\/1\/9\/605\/3405\/\/Ex.12.MCQ_Final%20File_html_m2f093f8e.gif\")
<\/span><\/span><\/p>\n <\/p>\n
Therefore the area of a triangle say A, <\/em>having sides 16 cm, 30 cm and 34 cm is given by<\/span><\/span><\/p>\na<\/em> = 16 cm ; b<\/em> = 30 cm ; c<\/em> = 34 cm<\/span><\/span><\/p>\n![\"\"](\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/iit_pretests\/1\/9\/605\/3405\/\/Ex.12.MCQ_Final%20File_html_m703ed80a.gif\")
<\/p>\n
<\/p>\n
<\/p>\n
<\/p>\n
Therefore the area of the triangle is <\/span><\/span><\/p>\n![\"\"](\"https:\/\/img-nm.mnimgs.com\/img\/study_content\/iit_pretests\/1\/9\/605\/3405\/\/Ex.12.MCQ_Final%20File_html_71ad7da9.gif\")
Hence, the correct option is (b).<\/p>\n
\n
<\/span>Question 2:<\/span><\/h4>\n<\/span>The base of an isosceles right triangle is 30 cm. Its area is
(a) 225 cm2<\/sup>
(b) 225 3<\/mn><\/msqrt><\/math>“>3<\/span><\/span>\u2013\u221a<\/span><\/span><\/span>3<\/span><\/span> cm2<\/sup>
(c) 225 2<\/mn><\/msqrt><\/math>“>2<\/span><\/span>\u2013\u221a<\/span><\/span><\/span>2<\/span><\/span> cm2<\/sup>
(d) 450 cm2<\/sup><\/span><\/h4>\n<\/div>\n\n
<\/span> <\/span><\/h4>\n<\/span>ANSWER:<\/span><\/h4>\nLet<\/mi> <\/mo>ABC<\/mi> <\/mo>be<\/mi> <\/mo>the<\/mi> <\/mo>right<\/mi> <\/mo>triangle<\/mi> <\/mo>in<\/mi> <\/mo>which<\/mi> <\/mo>∠<\/mo>B<\/mi>=<\/mo>90<\/mn>°<\/mo>.<\/mo>Now<\/mi>,<\/mo> <\/mo>base<\/mi>=<\/mo>BC<\/mi>;<\/mo> <\/mo>perpendicular<\/mi>=<\/mo>AB<\/mi>;<\/mo> <\/mo>Hypotenuse<\/mi>=<\/mo>AC<\/mi>Now<\/mi>,<\/mo> <\/mo>BC<\/mi>=<\/mo>30<\/mn> <\/mo>cm<\/mi> <\/mo>given<\/mi><\/mfenced>Now<\/mi>,<\/mo> <\/mo>∆<\/mo>ABC<\/mi> <\/mo>is<\/mi> <\/mo>an<\/mi> <\/mo>isosceles<\/mi> <\/mo>right<\/mi> <\/mo>angled<\/mi> <\/mo>∆<\/mo> <\/mo>and<\/mi> <\/mo>we<\/mi> <\/mo>know<\/mi> <\/mo>that<\/mi> <\/mo>hypotenuse<\/mi> <\/mo>is<\/mi> <\/mo>the<\/mi> <\/mo>longest<\/mi> <\/mo>side<\/mi> <\/mo>of<\/mi> <\/mo>the<\/mi> <\/mo>right<\/mi> <\/mo>∆<\/mo>.<\/mo>So<\/mi>,<\/mo> <\/mo>AB<\/mi>=<\/mo>BC<\/mi>=<\/mo>30<\/mn> <\/mo>cm<\/mi>area<\/mi> <\/mo>of<\/mi> <\/mo>∆<\/mo>ABC<\/mi>=<\/mo>
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-17-MCQS.jpg\")
<\/p>\n