- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.
Geometry and Mensuration: Level 3 Test 10
Congratulations - you have completed Geometry and Mensuration: Level 3 Test 10.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE: EB = 1: 2 and DF is perpendicular to MN such that NL: LM = 1: 2. The length of DH in cm is
$ \displaystyle 2\sqrt{2}-1$ | |
$ \displaystyle \frac{\left( 2\sqrt{2}-1 \right)}{2}$ | |
$ \displaystyle \frac{\left( 3\sqrt{2}-1 \right)}{2}$ | |
$ \displaystyle \frac{\left( 2\sqrt{2}-1 \right)}{3}$ |
Question 1 Explanation:
$ \begin{array}{l}MO=1.5\text{ }cm,\text{ }ON=1.5\text{ }cm\\AO=OB=1.5\text{ }cm.\\AE:EB=1:2,\text{ }so\text{ }AE=1cm\text{ }and\text{ }EB=2cm.\\So,\text{ }EO=0.5\text{ }cm.HL=0.5\text{ }cm\\Similarly,\text{ }OL=0.5\text{ }cm\text{ }and\text{ }EH=0.5\text{ }cm.\\Now,\text{ }OD=\text{ }1.5cm\text{ }and\text{ }thus\text{ }DL=\sqrt{{{1.5}^{2}}-{{0.5}^{2}}}=\sqrt{2}\\DH=DL-HL=~\frac{2\sqrt{2}-1}{2}\end{array}$
Question 2 |
Consider the triangle ABC shown in the following figure where BC= 12 cm, DB= 9 cm, CD= 6 cm and ∠BCD= ∠BAC.
What is the ratio of the perimeter of the triangle ADC to that of the triangle ΔBDC?
7/9 | |
8/9 | |
6/9 | |
5/9 |
Question 2 Explanation:
$ \begin{array}{l}Triangle\text{ }BDC\text{ }is\text{ }equivalent\text{ }to\text{ }triangle\text{ }BAC\\Therefore,\\\frac{BD}{BC}=\frac{CD}{AC}=\frac{BC}{AD}\\AC=\frac{12}{9}\times 6=8\\AD=\frac{12}{9}\times 12-9=16-9\\\frac{\Delta ADC}{\Delta BDC}=\frac{8+7+6}{9+12+6}=\frac{21}{27}=\frac{7}{9}\end{array}$
Question 3 |
P, Q, S and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?
$ \displaystyle 2\,r\left( 1+\sqrt{3} \right)\,$ | |
$ \displaystyle 2r\,\left( 2+\sqrt{3} \right)$ | |
$ \displaystyle r\,\left( 1+\sqrt{5} \right)$ | |
$ \displaystyle 2r+\sqrt{3}$ |
Question 3 Explanation:
$\begin{array}{l}Let\text{ }PQ=a\\From\text{ }the\text{ }equilateral\text{ }triangle\text{ }we\text{ }can\text{ }find\,a=r\sqrt{3}\\Now\text{ }QSR\text{ }is\text{ }an\text{ }isosceles\text{ }triangle\\QS=\sqrt{3{{(\frac{r}{2})}^{2}}+{{(\frac{r}{2})}^{2}}}=r\\Thus\text{ }the\text{ }total\text{ }perimeter\text{ }of\text{ }quadrilateral\text{ }PQSR\text{ }=2r+2r\sqrt{3}=2r(\sqrt{3}+1)\end{array}$
Question 4 |
Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side a in the following steps: the original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of hte smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is
$ \displaystyle \sqrt{2}:1$ | |
$ \displaystyle 1:1$ | |
$ \displaystyle \frac{n\,\left( 4-\pi \right)}{4n-\pi }$ | |
$ \displaystyle \frac{4n-\pi }{n\,\left( 4-\pi \right)}$ |
Question 4 Explanation:
The area of the scrap cloth generated in both case are proportional.
Thus the total amount of the cloth remaining will be same in both cases.
Thus the correct option is (a)
Question 5 |
In the adjoining figure, AC +AB = 5AD and AC - AD = 8. Then, the area of the rectangle ABCD is:
36 | |
50 | |
60 | |
Cannot be answered |
Question 5 Explanation:
$\displaystyle \begin{array}{l}AC+AB=5AD=5BC\\A{{C}^{2}}-A{{B}^{2}}=B{{C}^{2}}\\=>AC-AB=\frac{BC}{5}\\AC=\frac{13}{5}BC\\AC-AD=8\\=>AC-BC=8\\=>\frac{8BC}{5}=8\\=>BC=5\\AB=12(Pythagorean\,triplet)\\Area=12\times 5=60\end{array}$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
