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The figure below is not drawn to scale ABCD is a rhombus. CDF, BCG and EFG are straight line. Find$\angle$y.
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$\angle$BAC = $180^\circ - 56^\circ = 124^\circ$
$\angle$DCG = $180^\circ - 124^\circ = 56^\circ$
$\angle$DFG = $180^\circ - 56^\circ - 155^\circ = 69^\circ$
y = $180^\circ - 69^\circ = 111^\circ$
$\angle$y is 111$^\circ$
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$\angle$BAC = $180^\circ - 56^\circ = 124^\circ$
$\angle$DCG = $180^\circ - 124^\circ = 56^\circ$
$\angle$DFG = $180^\circ - 56^\circ - 155^\circ = 69^\circ$
y = $180^\circ - 69^\circ = 111^\circ$
$\angle$y is 111$^\circ$
In the figure below, not drawn to scale, ABCG is a rhombus. BGF, CDF and EDG are straight line, Find
(a) $\angle$x
(b) $\angle$y.
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(a) $\angle$DCG = $90^\circ - 68^\circ = 28^\circ$
$\angle$CGD = $180^\circ - 110^\circ - 28^\circ = 42^\circ$
$\angle$AGC = $360^\circ โ \frac{(68 \times2)}{2} = 112^\circ$
$\angle$x = $360^\circ - 112^\circ - 42^\circ = 206^\circ$ (b) $\angle$GBC = $112^\circ รท 2 = 56^\circ$
$\angle$y = $180^\circ - 56^\circ -96^\circ = 28^\circ$
In the figure below, ABCD is a rhombus and CDE is an isosceles triangle. $\angle$CDE = 36$^\circ$ and BCE is a straight line. Find$\angle$x.
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In the figure below, STUV is a parallelogram and VWXS is a rhombus. Which of the following pairs of line are parallel?
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In the figure below. ABCD is a rhombus, $\angle$ABC =102$^\circ $ and$\angle$CAE=19$^\circ $. Find $\angle$EAB.
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20$^\circ$ $\rightarrow$ $180^\circ โ 120^\circ = 78^\circ$
$78^\circ รท 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
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20$^\circ$ $\rightarrow$ $180^\circ โ 120^\circ = 78^\circ$
$78^\circ รท 2 = 39^\circ$
$39^\circ - 19^\circ = 20^\circ$
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In the figure, ABCD and EFGH are identical rhombuses and BKMN is a trapezium. Which one of the following statements is true?
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The figure below is not drawn to scale. ABCD is a rhombus and $\angle$BCF is 98$^\circ$. BDF, CGF, EGD are straight line. Find $\angle$DFG.
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$\angle$BDC $\rightarrow$ $(180^\circ - 68^\circ) \div 2 = 56^\circ$ = $\angle$FBC
$\angle$DFG $\rightarrow$ $180^\circ - 56^\circ - 98^\circ = 26^\circ$
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$\angle$BDC $\rightarrow$ $(180^\circ - 68^\circ) \div 2 = 56^\circ$ = $\angle$FBC
$\angle$DFG $\rightarrow$ $180^\circ - 56^\circ - 98^\circ = 26^\circ$
In the diagram below, ABCD is a rhombus. E is the midโpoint BC and AC. Find the area the shaded part.
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In the figure, ABCD is a rhombus. EDF is a straight line $\angle$BAD = 58$^\circ$, and $\angle$ADE = 78$^\circ$. Find $\angle$FDC.
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In the diagram below, ABCO and FODE are identical rhombuses and AOF and OCD are identical equilateral triangle. What fraction of the figure ABCDEF is shaded?
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In the figure below, not drawn to scale, ABCD is a rhombus. Given that BCE is a straight line, Find $\angle$p + $\angle$q.
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The figure not drawn to scale, is made up of a triangle ABC and two rhombuses, PSRQ and PTCU. $\angle$ TPU = 79$^\circ$. Find $\angle$ d + $\angle$ e + $\angle$ f + $\angle$ g.
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$\angle$d + $\angle$e = $180^\circ = 79^\circ = 101^\circ$
$\angle$d + $\angle$e + $\angle$f + $\angle$g = $101^\circ + 202^\circ = 303^\circ$
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$\angle$d + $\angle$e = $180^\circ = 79^\circ = 101^\circ$
$\angle$d + $\angle$e + $\angle$f + $\angle$g = $101^\circ + 202^\circ = 303^\circ$
In the figure below, ABCD is a rhombus. $\angle$EDB = 97$^\circ$, $\angle$FEG = 110$^\circ$,and $\angle$BAD = 66$^\circ$. Give, that GED, FED and FDB are straight lines, Find$\angle$FCD,
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$\angle$CBD $\rightarrow$ $180^\circ โ 66^\circ \div 2 = 57^\circ$
$\angle$EDC $\rightarrow$ $97^\circ โ 57^\circ = 40^\circ$
$\angle$FCD $\rightarrow$ $180^\circ โ 40^\circ โ 110^\circ = 30^\circ$
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$\angle$CBD $\rightarrow$ $180^\circ โ 66^\circ \div 2 = 57^\circ$
$\angle$EDC $\rightarrow$ $97^\circ โ 57^\circ = 40^\circ$
$\angle$FCD $\rightarrow$ $180^\circ โ 40^\circ โ 110^\circ = 30^\circ$
The figure below is not drawn to scale. ABCD is a rhombus. DBE is a straight line. BC is parallel to EF. Find $\angle$DEF.
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In the figuare, ABCD is a rhombus. ACD, AHF and DEF are straight line. HE is parallel to GF and HE = HF. $\angle$ABC = 110$^\circ$ and $\angle$HFG = 20$^\circ$. Find $\angle$CDE.
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$\angle$EHF = $\angle$GFH = 20$^\circ$
$\angle$HEF = $\angle$HFE = $(180 - 20) \div 2 = 80^\circ$
$\angle$CAH = $\angle$HCA = $(180 - 110) \div 2 = 35^\circ$
$\angle$CDE = $180 โ 80 โ 35 = 65^\circ$
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$\angle$EHF = $\angle$GFH = 20$^\circ$
$\angle$HEF = $\angle$HFE = $(180 - 20) \div 2 = 80^\circ$
$\angle$CAH = $\angle$HCA = $(180 - 110) \div 2 = 35^\circ$
$\angle$CDE = $180 โ 80 โ 35 = 65^\circ$
The figure below shows a rhombus divided into 6 parts with 3 straight lines. Find the sum of $\angle$ X and $\angle$ Y.
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$360 โ 274 = 86$
$(360 โ 86 - 86) รท 2 = 94$
$(360 โ 102 - 102) รท 2 = 47$
$360 โ 102 โ 47 = 221$
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$360 โ 274 = 86$
$(360 โ 86 - 86) รท 2 = 94$
$(360 โ 102 - 102) รท 2 = 47$
$360 โ 102 โ 47 = 221$
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$5 + 1 = 6$
$6 \times 6 = 36$
$36 โ 25 = 11$
$5 \times 5 = 25$
$\frac{11}{25} \times 100 = 44%$
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$5 + 1 = 6$
$6 \times 6 = 36$
$36 โ 25 = 11$
$5 \times 5 = 25$
$\frac{11}{25} \times 100 = 44%$
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$180^\circ - 74^\circ = 106^\circ$
$(180^\circ - 74^\circ) \div 2 = 53^\circ$
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$180^\circ - 74^\circ = 106^\circ$
$(180^\circ - 74^\circ) \div 2 = 53^\circ$
The figure below is made up of a right โangled triangle and a rhombus overlapping each other. $\frac{2}{9}$ of the triangle and $\frac{1}{2}$ of the rhombus are unshaded. The base and height of the triangle are 18cm and 28cm respectively. What is the area of the rhombus?
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392cm$^2$ $\rightarrow$ 9u $\rightarrow$ $\frac{1}{2} \times 18cm \times 28cm = 252cm^2$
1U $\rightarrow$ $252cm รท 9 = 28cm^2$
14u $\rightarrow$ $28cm^2 \times 14 = 392cm^2$
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392cm$^2$ $\rightarrow$ 9u $\rightarrow$ $\frac{1}{2} \times 18cm \times 28cm = 252cm^2$
1U $\rightarrow$ $252cm รท 9 = 28cm^2$
14u $\rightarrow$ $28cm^2 \times 14 = 392cm^2$
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