Question
Evaluate the following :$1.81 \times 1.81 - 1.81 \times 2.19 + 2.19 \times 2.19$
✓
Answer
$1.81 \times 1.81-1.81 \times 2.19+2.19 \times 2.19$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)+(1.81 \times 2.19)$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)$
$=(1.81-2.19)^2+(2.00-0.19)^2+(1.81 \times 2.19)$
$=(0.38)^2+\left(4-(0.19)^2\right)$
$= 0.1444+(4-0.0361)$
$= 0.1444+3.9639$
$= 4.1083$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)+(1.81 \times 2.19)$
$=(1.81)^2-(1.81 \times 2.19)+(2.19)^2-(1.81 \times 2.19)$
$=(1.81-2.19)^2+(2.00-0.19)^2+(1.81 \times 2.19)$
$=(0.38)^2+\left(4-(0.19)^2\right)$
$= 0.1444+(4-0.0361)$
$= 0.1444+3.9639$
$= 4.1083$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Draw the graph ofequation $\frac{x}{4}+\frac{y}{5}=1$ Use the graph drawn to find:$(i) x_1$, the value of $x$, when $y = 10(ii) y_1$, the value of $y$, when $x = 8$.
→In the given figure, $AB \| DC \| EF , AD \| BE$ and $DE \| AF$. Prove that : ar $(\| \operatorname{gm} DEFH )=\operatorname{ar}(\| gm ABCD )$.
→Find the area of quadrilateral ABCD in which
AB = 29 cm, BC = 21 cm, AC=20 cm CD = 34 cm and DA = 42 cm.
AB = 29 cm, BC = 21 cm, AC=20 cm CD = 34 cm and DA = 42 cm.
If $2^{\mathrm{x}}=4^{\mathrm{y}}=8^{\mathrm{z}}$ and $\frac{1}{2 x}+\frac{1}{4 y}+\frac{1}{8 z}=4$, find the value of $\mathrm{x}$.
→Simplify by rationalising the denominator in the following.$\frac{2}{3+\sqrt{7}}$
→In the adjoining figure, $AB = AC$. If $DB \perp BC$ and EC $\perp BC$, prove that :
(i) $BD = CE$
(ii) $AD = AE$.
→(i) $BD = CE$
(ii) $AD = AE$.
Use graph paper for this question. Draw the graph of $2x - y - 1 = 0$ and $2x + y = 9$ on the same axes. Use $2 \ cm = 1$ unit on both axes and plot only $3$ points per line. Write down the coordinates of the point of intersection of the two lines.
→Calculate the amount and the compound interest on ₹ 64000 for 3 years at $7 \frac{1}{2} \%$ per annum compounded annually.
→In a $\triangle A B C, A C>A B, D$ is the midpoint $B C$, and $A E \perp B C$. Prove that: $A B^2=A D^2-B C \times C E+$
$\frac{1}{4} BC ^2$
→$\frac{1}{4} BC ^2$
Express the following in terms of $\log 2$ and $\log 3: \log \frac{225}{16}-2 \log \frac{5}{9}+\log \left(\frac{2}{3}\right)^5$
→