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## Geometry and Mensuration: Level 3 Test 4

Congratulations - you have completed Geometry and Mensuration: Level 3 Test 4. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
 Question 1
$\displaystyle \begin{array}{l}If\,a,b,c\,\,are\,\,the\,\,sides\,\,of\,\,a\,\,triangle\,\,and\,\,{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ba+ca+ab\\then\text{ }the\text{ }triangle\text{ }is\end{array}$
 A Equilateral B Isosceles C Right-angled D Obtuse-angled
Question 1 Explanation:
$\displaystyle \begin{array}{l}{{a}^{2}}+{{b}^{2}}+{{c}^{2}}=ba+ca+ab\\is\text{ }an\text{ }identity\text{ }when\text{ }a=b=c\\Thus\text{ }the\text{ }correct\text{ }option\text{ }is\text{ }\left( a \right)\end{array}$
 Question 2
In the given diagram, ABCD is a rectangle with AE=EF =FB. What is the ratio of the area of the triangle CEF and that of the rectangle?
 A 1: 4 B 1: 6 C 2: 5 D 2:3
Question 2 Explanation:

Since EF = 1/3 of AB
CEF is 1/3 of CAB
= 1/3 x ½ x ABCO
= 1/6 of ABCO
 Question 3
In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC= AC and BF= BC. Angle D= 40 degrees then what is angle ACB in degrees?
 A 140 B 70 C 100 D None of these
Question 3 Explanation:

Let the angle AEC = x and BFC = y,
AEC=CAE = x and ACE = 180-2x
Similarly,
BCF = 180-2y.
Thus ACB = 180-{360-(2x+2y)} =2(x+y)-180
Now x+y = 180-40 =140
ACB = 100.
Correct option is (c).
 Question 4
In the given figure, ACB is a right angled triangle, CD is the altitude. Circles are inscribed within the triangle ACD, BCD. P and Q are the centres of the circles. The distance PQ is?
 A 5 B √50 C 7 D 8
Question 4 Explanation:

$\begin{array}{l}Now\,\,{{15}^{2}}-{{x}^{2}}={{20}^{2}}-\left( 25-x \right){}^{2}\\x=9cm\\Thus\text{ }area\text{ }of\text{ }the\text{ }triangle~ADC=\frac{1}{2}\times 9\times \sqrt{{{15}^{2}}-{{9}^{2}}}=54\,c{{m}^{2}}\\The\text{ }area\text{ }of\text{ }the\text{ }triangle\text{ }BCD=\frac{1}{2}\times 12\times 16=96\,c{{m}^{2}}\\The\text{ }radius\text{ }of\text{ }the\text{ }circle\text{ }inside\text{ }ADC=\frac{2\left( \frac{1}{2}\times 9\times 12 \right)}{15+9+12}=3cm\\The\text{ }radius\text{ }of\text{ }the\text{ }circle\text{ }inside\text{ }BCD=4cm\\Thus\text{ }one\text{ }radius\text{ }is\text{ }bigger\text{ }than\text{ }the\text{ }other.\\\begin{array}{*{35}{l}} Therefore\text{ }the\text{ }length\text{ }of\text{ }PQ\text{ }=\text{ }7\text{ }cm \\ Q{{Q}^{''}}~=1\text{ }cm \\ \begin{array}{l}Therefore\text{ }the\text{ }required\text{ }distance\text{ }=\sqrt{{{7}^{2}}+{{1}^{2}}}\\=\sqrt{50}\end{array} \\ \end{array}\end{array}$
 Question 5
In the figure below, ABCD is a rectangle. The area of the isosceles right triangle ABE -7 cm2; EC= 3 (BE). The area of ABCD (in cm22) is:
 A 21 B 28 C 42 D 56
Question 5 Explanation:

AB x BE = AB x AB=AE=AE = 2x7.